The best length condition you'll ever see

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3564 posts

#1 h Apr 17, 2019, 10:59 PM • 36 Y

Y by trumpeter, Reef334, wu2481632, fatant, Ultroid999OCPN, sriraamster, Vietjung, Wizard_32, pggp, amar_04, rashah76, OlympusHero, myh2910, Muaaz.SY, icematrix2, ImSh95, mijail, MiraclesINmaths, Supercali, megarnie, khina, HWenslawski, suvamkonar, Pitagar, jhu08, centslordm, Lamboreghini, aopsuser305, rayfish, mathleticguyyy, sabkx, Adventure10, juicetin.kim, aidan0626, Rounak_iitr, Funcshun840

Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$ . The diagonals of $ABCD$ intersect at $E$ . Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$ . Show that line $PE$ bisects $\overline{CD}$ .

Proposed by Ankan Bhattacharya

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yayups

1614 posts

yayups

#2 h Apr 17, 2019, 11:00 PM • 12 Y

Y by a000, pad, sioly, anantmudgal09, Ravana, ImSh95, megarnie, EricShi1685, suvamkonar, Aspiring_Mathletes, Adventure10, Mango247

https://lh3.googleusercontent.com/-RihqykDRN94/XLeuaX_tJ9I/AAAAAAAAFHY/MMul-UoqMZ8sB3RJFjlto5dwAouRTihSACK8BGAs/s0/2019-04-17.png

It's easy to see by monotonicty that there is only one point satisfying $\angle APD=\angle BPC$ . Thus, we may redefine $P=ME\cap AB$ . We'll show that $\angle APD=\angle BPC$ .

Since $AD^2+BC^2=AB^2$ , we may choose $X\in(AB)$ so that $AX=AD$ and $BX=BC$ . The first claim is that $P$ is the foot from $X$ to $AB$ . Redefine $P$ in this manner, we'll show that $P,E,M$ collinear. Note that
$\frac{AP}{BP}=\frac{AX\cos\angle XAB}{BX\sin\angle XAB}=\frac{AX^2}{BX^2}.$ But
$\frac{AX}{BX}=\frac{AD}{BC}=\frac{2R\sin\angle ABD}{2R\sin\angle BAC}=\frac{AE}{BE},$ where the last step is law of sines on $AEB$ . Thus,
$\frac{AP}{BP}=\frac{AE^2}{BE^2},$ so $EP$ is the $E$ -symmedian of $AEB$ . There are many ways to see that this implies that $P,E,M$ collinear, but perhaps the easiest is to invert at $E$ . Let $E'\in(AEB)$ so that $(EE';AB)=-1$ (read: the harmonic conjugate of $E$ in $AB$ in the complex projective line). Clearly, $E,P,E'$ collinear. Let $\phi$ be inversion at $E$ swapping $(A,C)$ and $(B,D)$ . Note that $\phi((ABE))=CD$ , and
$-1=\phi((EE';AB))=(\phi(E')\infty,CD),$ so $\phi(E')=M$ , so $E',E,M$ collinear, so $E,P,M$ collinear, as desired.

We'll now show that $\angle APD=\angle BPC$ . To do this, let $D'=DP\cap(ABCD)$ and $C'=CP\cap(ABCD)$ . We claim that $AD=AD'$ and $BC=BC'$ . We'll show the former, and the latter follows by symmetry. Let $\omega=(ABCD)$ and $\omega_A$ the circle centered at $A$ with radius $AD$ . The power of $P$ with respect to $\omega_A$ is
$AP^2-AD^2=AP^2-AX^2=-PX^2,$ and the power of $P$ with respect to $\omega$ is $-|PA|\cdot|PB|=-PX^2$ , so $P$ is on the radical axis of $\omega_A$ and $\omega$ . Thus, the radical axis is $DP$ , so $D'$ is on the radical axis. But since $D'\in\omega$ , we must also then have that $D'\in\omega_A$ , so $AD=AD'$ , as desired.

The problem is clearly equivalent now to $AB$ being the external angle bisector of $CC'$ and $DD'$ . But $DD'\parallel\ell_A$ (the tangent to $\omega$ at $A$ ), and $CC'\parallel\ell_B$ (the tangent to $\omega$ at $B$ ), so the external bisector of $CC',DD'$ is parallel to the external bisector of $\ell_A$ and $\ell_B$ . But the external bisector of those two tangents is clearly parallel to $AB$ , and as $AB,CC',DD'$ are concurrent, it's clear that $AB$ is the external bisector of $CC'$ and $DD'$ . $\blacksquare$

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trumpeter

3332 posts

trumpeter

#3 h Apr 17, 2019, 11:00 PM • 10 Y

Y by SHREYAS333, Arshia.esl, to_chicken, Jalil_Huseynov, sabkx, Adventure10, Mango247, Begli_I., pineapply, cowcow

If $M$ is the midpoint of $CD$ , let $Q$ be where $EM$ and $AB$ meet. The goal is to show that $P=Q$ . Note that since $P$ is unique (try moving $P$ along segment $AB$ ), it suffices to show $\angle{AQD}=\angle{BQC}$ .

Observe that since $M,E,Q$ collinear, $\text{[math]}$ and $EM$ are corresponding isogonal lines in similar triangles $ECD$ and $EAB$ . So $\text{[math]}$ is a symmedian and thus $\frac{AQ}{QB}=\frac{AE^2}{EB^2}$ . But by the Law of Sines, $\frac{AE}{EB}=\frac{\sin\angle{ABE}}{\sin\angle{EAB}}=\frac{\sin\angle{ABD}}{\sin\angle{CAB}}=\frac{AD}{BC}$ so $\frac{AQ}{QB}=\frac{AD^2}{BC^2}$ . Since $AQ+QB=AB$ , it follows that $AQ=\frac{AD^2}{AB}$ . Then $AD$ is tangent to $(BQD)$ so $\angle{QDA}=\angle{ABD}$ . Then $\angle{AQD}=\pi-\angle{QDA}-\angle{DAQ}=\pi-\angle{ABD}-\angle{DAB}=\angle{BDA}.$ Similarly, $\angle{BQC}=\angle{ACB}$ . But $\angle{BDA}=\angle{ACB}$ by cyclic quad, so $\angle{AQD}=\angle{BQC}$ as desired.

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tworigami

844 posts

tworigami

#4 h Apr 17, 2019, 11:01 PM • 4 Y

Y by Justpassingby, Aspiring_Mathletes, Adventure10, Mango247

Let $\omega_A$ denote the circle with center $A$ and radius $AD$ . Let $\omega_B$ denote the circle with center $B$ and radius $BC$ . We claim that these two circles necessarily intersect.

Proof: Assume for the sake of contradiction that they don't. Then $AD + BC < BC \implies AD^2 + BC^2 + 2AD \cdot BC < AB^2 \implies 2AD \cdot BC < 0$ which is a contradiction.

Hence, they intersect at a point, say $X$ . Then by the Pythagorean theorem, $\angle AXB = 90^\circ$ so $\omega_A$ and $\omega_B$ are orthogonal. Let $P'$ be the projection of $X$ onto $\overline{AB}$ . Upon inverting about $\omega_A$ we see that $P'D$ and $AC$ meet on $\omega_B$ at $M$ and by symmetry, $P'C$ and $BD$ meet on $\omega_A$ at $N$ . Since $B$ is mapped to $P'$ and $M$ is mapped to $C$ , $AXYB$ is a cyclic quadrilateral to $\angle AND = \angle BMC$ hence $\angle AP'D = \angle BP'C$ where we have used the fact that quadrilaterals $ABP'N$ and $BCP'M$ are cyclic.

Moreover, since $XY$ and $CD$ are both antiparallel to $AB$ they are parallel so by Ceva's theorem $P'E$ passes through the midpoint of $BC$ .

It remains to show that $P' = P$ . Clearly, $P'$ satisfies the angle condition for $P$ so we need to prove that $P$ is unique. To do this, we show a unique construction for $P$ . Let the perpendicular bisector of $CD$ meet $AB$ at $T$ and then the circumcircle of $\triangle TCD$ meet $AB$ again at $P$ .

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62861

3564 posts

62861

#5 h Apr 17, 2019, 11:01 PM • 15 Y

Y by fatant, v_Enhance, anantmudgal09, Centralorbit, Vietjung, Kitkat2207, amar_04, myh2910, tigerzhang, suvamkonar, Lamboreghini, aopsuser305, megarnie, Adventure10, Mango247

Here was the original solution I found, with some suggestions from Evan to make it simpler.

Solution

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jj_ca888

2726 posts

jj_ca888

#6 h Apr 17, 2019, 11:03 PM • 2 Y

Y by Adventure10, Mango247

SYMMEDIAN!!!!! :rage:

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budu

1515 posts

budu

#7 h Apr 17, 2019, 11:04 PM • 9 Y

Y by expiLnCalc, sriraamster, kevinmathz, centslordm, sabkx, Adventure10, Mango247, Mango247, Mango247

Solution

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PARISsaintGERMAIN

246 posts

PARISsaintGERMAIN

#8 h Apr 17, 2019, 11:04 PM • 4 Y

Y by Phie11, HoRI_DA_GRe8, Adventure10, Rounak_iitr

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -30.10774683338543, xmax = 20.292988546313076, ymin = -14.748881041259867, ymax = 20.17411985764882; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0,1); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-5.21,0.62), 6.5651208575055), linewidth(0.4) + xfqqff); draw(circle((-11.773257257435112,0.4635837400426075), 7.761816012333698), linewidth(0.8) + qqffff); draw(circle((1.3532572574351138,0.7764162599573926), 10.590441902267118), linewidth(0.8) + qqffff); draw((-7.335393968675698,6.8315627946250945)--(-7.037079805882728,-5.685758578996604), linewidth(0.4) + wrwrwr); draw(circle((-5.371985513785333,7.41694425145151), 9.451216976541685), linewidth(0.8) + red); draw((-14.821393813947864,7.601836421348473)--(3.2945646862479396,11.187409735708462), linewidth(0.8) + wrwrwr); draw((-14.821393813947864,7.601836421348473)--(-11.773257257435112,0.4635837400426075), linewidth(0.8) + wrwrwr); draw((1.3532572574351138,0.7764162599573926)--(3.2945646862479396,11.187409735708462), linewidth(0.8) + wrwrwr); draw((-14.821393813947864,7.601836421348473)--(-4.407768443334378,-1.984959161087927), linewidth(0.8) + wrwrwr); draw((-14.821393813947864,7.601836421348473)--(1.3532572574351138,0.7764162599573926), linewidth(0.8) + wrwrwr); draw((3.2945646862479396,11.187409735708462)--(-11.773257257435112,0.4635837400426075), linewidth(0.8) + wrwrwr); draw((-11.773257257435112,0.4635837400426075)--(1.3532572574351138,0.7764162599573926), linewidth(0.8) + wrwrwr); draw((-9.032292019977847,-1.2967035850400774)--(3.2945646862479396,11.187409735708462), linewidth(0.8) + wrwrwr); draw((-7.186236887279213,0.5729021078142449)--(-5.763414563849959,9.394623078528468), linewidth(0.8) + wrwrwr); draw((-9.119258911737576,2.352444423319879)--(-4.532772070486577,3.2602178002833475), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((1.3532572574351138,0.7764162599573926),linewidth(4pt) + dotstyle); label("$B$", (1.8941041775996061,0.03126540659287662), NE * labelscalefactor); dot((-11.773257257435112,0.4635837400426075),linewidth(4pt) + dotstyle); label("$A$", (-12.624446757901781,-0.05593309752424867), NE * labelscalefactor); dot((-7.335393968675698,6.8315627946250945),linewidth(4pt) + dotstyle); label("$X$", (-7.17454025058144,7.18154274419715), NE * labelscalefactor); dot((-7.186236887279213,0.5729021078142449),linewidth(4pt) + dotstyle); label("$P$", (-7.000143242347189,-0.6663226263441258), NE * labelscalefactor); dot((3.2945646862479396,11.187409735708462),linewidth(4pt) + dotstyle); label("$C$", (3.463677251707864,11.541467950053415), NE * labelscalefactor); dot((-14.821393813947864,7.601836421348473),linewidth(4pt) + dotstyle); label("$D$", (-15.589195897884046,7.8791307771341526), NE * labelscalefactor); dot((-4.407768443334378,-1.984959161087927),linewidth(4pt) + dotstyle); label("$G$", (-4.166191858540612,-3.0206822375065085), NE * labelscalefactor); dot((-9.032292019977847,-1.2967035850400774),linewidth(4pt) + dotstyle); label("$H$", (-9.790495374095203,-2.06149869221813), NE * labelscalefactor); dot((-6.611342353611276,4.137338144623156),linewidth(4pt) + dotstyle); label("$E$", (-7.218139502640003,4.739984628917642), NE * labelscalefactor); dot((-9.119258911737576,2.352444423319879),linewidth(4pt) + dotstyle); label("$L$", (-9.659697617919516,3.0396137986336993), NE * labelscalefactor); dot((-4.532772070486577,3.2602178002833475),linewidth(4pt) + dotstyle); label("$N$", (-3.8173978420721104,2.9960145465751364), NE * labelscalefactor); dot((-7.037079805882728,-5.685758578996604),linewidth(4pt) + dotstyle); label("$Y$", (-7.305338006757128,-6.6830194104257705), NE * labelscalefactor); dot((-5.763414563849959,9.394623078528468),linewidth(4pt) + dotstyle); label("$M$", (-5.604967176473182,9.753898615652346), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $\Gamma_1$ and $\Gamma_2$ be the circle centered at $A, B$ with radius $AD, BC$ respectively.
Let $X,Y$ be the intersection of these two circles.
Notice that $X, Y$ lies on the circle with diameter $AB$ by Pythagorean theorem.
Let $P'$ be the intersection of $AB$ and $XY.$
We claim that $P=P'.$
First, notice that $XY$ is the radical axis of circle $\Gamma_1$ and $\Gamma_2.$
Since $AB$ is the line connecting the center, we have $XY\perp AB.$
Let $G=DP'\cap \Gamma_1$ and $H=CP'\cap\Gamma_2.$
By the power of a point, $DP'\times GP'=AX^2 -AP'^2=P'X^2=AP'\times BP'$ that $(A,D,B,G)$ cyclic. Similarly, we have $(A,C,B,H)$ cyclic.
Therfore, points $G,H$ lies on the circle $ABCD.$
Now, $AD^2=AX^2 =AP'\times AB,$ that $\triangle ABD$ and $\triangle ADP'$ is similar. Thus, $\angle AP'D=\angle ADB.$
In the same way, we get $\angle CP'B=\angle ACB.$
Therefore, $\angle AP'D=\angle BP'C.$
Such point on the line $AB$ is unique(requires proof) so we have $P=P'.$

This implies that when we extend $DP$ and take the intersection with $ABCD$ which is $G$ in this case, we have $AG=AD$ and same for $H.$

Let $L,N$ be the intersection of $CA,DP$ and $BD,CP$ respectively.

Notice that $AD=AG$ and $BC=BH.$
By angle chase, we have $\angle LAB=\angle HAB$ and $\angle NBA=\angle GBA.$
Therefore, $L,N$ is symmetric with $H,G$ respect to line $AB.$
Since $(A,B,G,H)$ is cyclic, $(A,B,N,L)$ is cyclic as well.
Then, $\angle DCA=\angle DBA=\angle CLN$ that $CD\parallel LN.$
Let $M$ be the intersection of $PE$ and $CD.$
By Ceva's theorem, we have that $CM=DM.$

This post has been edited 2 times. Last edited by PARISsaintGERMAIN, Apr 18, 2019, 12:05 AM

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jj_ca888

2726 posts

jj_ca888

#9 h Apr 17, 2019, 11:04 PM • 1 Y

Y by Adventure10

how any points would I get if my solution would be complete if I proved PE was symmedian, but I didnt prove it

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PARISsaintGERMAIN

246 posts

PARISsaintGERMAIN

#10 h Apr 17, 2019, 11:15 PM • 1 Y

Y by Adventure10

wutt ignore

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numbersandnumbers

258 posts

numbersandnumbers

#11 h Apr 17, 2019, 11:18 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of $CD$ and define $P' = ME \cap AB$ ; by monotonicity, it suffices to show that $\angle AP'D = \angle BP'C$ .

Since $\triangle ABE \sim \triangle DCE$ , $EP'$ is the symmedian of $\triangle ABE$ . Therefore
$AP' = \frac{AB \cdot AE^2}{AE^2 + EB^2} = \frac{AB \cdot AD^2}{AD^2 + BC^2}.$ The last step is since $\triangle EAD \sim \triangle EBC$ . By the length condition, we then have $AP' = \frac{AD^2}{AB}$ . But then
$\frac{AP'}{AD} = \frac{AD}{AB},$ so $\triangle AP'D \sim \triangle ADB$ . Similarly, $\triangle BP'C \sim \triangle BCA$ . So
$\angle AP'D = \angle ADB = \angle ACB = \angle BP'C.$

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alifenix-

1547 posts

alifenix-

#12 h Apr 17, 2019, 11:23 PM • 38 Y

Y by wu2481632, yayups, 62861, Einstein314, Ultroid999OCPN, sriraamster, pad, SHARKYKESA, mira74, pi31415926535897, TLP.39, matharcher, no_name, Mudkipswims42, couplefire, FISHMJ25, Unstupid, tapir1729, kc5170, OlympusHero, myh2910, v4913, tigerzhang, 554183, Geometry285, Bedwarspro, megarnie, math31415926535, Iora, IMUKAT, Creeper1612, sabkx, jrsbr, kamatadu, Adventure10, aidan0626, vrondoS, Alex-131

thanks evan

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TheUltimate123

1739 posts

TheUltimate123

#13 h Apr 17, 2019, 11:23 PM • 3 Y

Y by FedeX333X, Adventure10, Rounak_iitr

Let $\Omega$ be the circumcircle of $ABCD$ , let $\Gamma$ be the circle with diameter $AB$ , and let $\omega_A$ , $\omega_B$ be the circles centered at $A$ , $B$ with radii $AD$ , $BC$ , respectively.

$[asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=purple+pink; pen qua=purple; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,Y,D,P,EE,M,Z,T,U,V; O=(0,0); A=dir(148); B=reflect( (0,0),(0,1))*A; C=dir(-70); path omega=circle(O,1); path gamma=circle((A+B)/2, length(A-B)/2); path wb=circle(B,length(C-B)); Y=intersectionpoint(gamma,wb); path wa=circle(A,length(Y-A)); D=intersectionpoints(omega,wa)[1]; P=extension(A,B,C,reflect(A,B)*D); EE=extension(A,C,B,D); M=(C+D)/2; Z=reflect(A,B)*Y; T=2*A*B/(A+B); U=2*foot(O,D,P)-D; V=2*foot(O,C,P)-C; filldraw(gamma,tfil,tri+dashed); draw(A--T--B,qua); draw(D--U,tri); draw(C--V,tri); filldraw(wa,sfil,sec+dashed); filldraw(wb,sfil,sec+dashed); draw(A--C,sec); draw(B--D,sec); filldraw(omega,fil,pri); draw(P--M,pri+Dotted); draw(A--B--C--D--cycle,pri); dot("$A$",A,dir(165)); dot("$B$",B,dir(15)); dot("$C$",C,S); dot("$D$",D,SW); dot("$P$",P,N); dot("$E$",EE,dir(60)); dot("$Y$",Y,dir(105)); dot("$Z$",Z,dir(265)); dot("$T$",T,N); dot("$U$",U,dir(75)); dot("$V$",V,dir(150)); dot(M); [/asy]$

The condition $AD^2+BC^2=AB^2$ means $\omega_A$ , $\omega_B$ are orthogonal, so their intersection points $Y$ , $Z$ lie on $\Gamma$ . Let $P$ be the midpoint of $\overline{YZ}$ . Then $P$ lies on $\overline{AB}$ , the radical axis of $\Omega$ , $\Gamma$ , so $P$ is the common radical center of $\Omega$ , $\Gamma$ , $\omega_A$ , $\omega_B$ .

Let the tangents to $\Omega$ at $A$ , $B$ meet at $T$ , and let $\overline{DP}$ , $\overline{CP}$ meet $\Omega$ again at $U$ , $V$ . Since $\overline{DU}$ is the radical axis of $\Omega$ , $\omega_A$ , we have $\overline{AT}\parallel\overline{DU}$ . Similarly $\overline{BT}\parallel\overline{CV}$ , so $\measuredangle APD=\measuredangle BAT=\measuredangle TBA=\measuredangle CPB$ . By monotonicity, $P$ is the point described in the problem statement.

Finally we have $\frac{PA}{PB}=\left(\frac{YA}{YB}\right)^2=\left(\frac{AD}{BC}\right)^2=\left(\frac{EA}{EB}\right)^2,$ so $\overline{EP}$ is the $E$ -symmedian of $\triangle EAB$ , which bisects $\overline{CD}$ .

This post has been edited 3 times. Last edited by TheUltimate123, Apr 29, 2020, 12:32 AM

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v_Enhance

6866 posts

v_Enhance

#14 h Apr 17, 2019, 11:26 PM • 17 Y

Y by yayups, TheUltimate123, Wizard_32, Omeredip, thczarif, myh2910, v4913, HamstPan38825, ChromeRaptor777, centslordm, Lamboreghini, sabkx, Adventure10, Mango247, Rounak_iitr, os31415, everythingpi3141592

Here is a solution by inversion.

By hypothesis, the circle $\omega_a$ centered at $A$ with radius $AD$ is orthogonal to the circle $\omega_b$ centered at $B$ with radius $BC$ . For brevity, we let $\mathbf{I}_a$ and $\mathbf{I}_b$ denote inversion with respect to $\omega_a$ and $\omega_b$ . We let $P$ denote the intersection of $\overline{AB}$ with the radical axis of $\omega_a$ and $\omega_b$ ; hence $P = \mathbf{I}_a(B) = \mathbf{I}_b(A)$ . This already implies that $\measuredangle DPA \overset{\mathbf{I}_a}{=} \measuredangle ADB = \measuredangle ACB \overset{\mathbf{I}_b}{=} \measuredangle BPC$ so $P$ satisfies the angle condition.

$[asy] size(10cm); pair A = dir(200); pair B = -conj(A); pair D = dir(140); pair O = origin; pair P = extension(D, foot(D, A, O), A, B); pair Z = 100*foot(P,O,B)-99*P; pair C = IP(P--Z, unitcircle); filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--D--cycle, deepcyan); filldraw(CP(A, D), invisible, red); filldraw(CP(B, C), invisible, orange); pair X = IP(CP(A, D), CP(B, C)); pair Y = OP(CP(A, D), CP(B, C)); draw(X--Y, yellow); draw(A--C, deepcyan+dotted); draw(B--D, deepcyan+dotted); pair K = extension(D, P, A, C); pair L = extension(C, P, D, B); draw(D--P, red+dashed); draw(C--P, orange+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$P$", P, dir(P)); dot("$C$", C, dir(C)); dot(X); dot(Y); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); /* TSQ Source: !size(12cm); A = dir 200 B = -conj(A) D = dir 140 O := origin P = extension D foot D A O A B Z := 100*foot(P,O,B)-99*P C = IP P--Z unitcircle unitcircle 0.1 palecyan / deepcyan A--B--C--D--cycle deepcyan CP A D 0.1 lightred / red CP B C 0.1 orange / orange X .= IP CP A D CP B C Y .= OP CP A D CP B C X--Y yellow A--C deepcyan dotted B--D deepcyan dotted K = extension D P A C L = extension C P D B D--P red dashed C--P orange dashed */ [/asy]$

Claim: The point $K = \mathbf{I}_a(C)$ lies on $\omega_b$ and $\overline{DP}$ . Similarly $L = \mathbf{I}_b(D)$ lies on $\omega_a$ and $\overline{CP}$ .

Proof. The first assertion follows from the fact that $\omega_b$ is orthogonal to $\omega_a$ . For the other, since $(BCD)$ passes through $A$ , it follows $P = \mathbf{I}_a(B)$ , $K = \mathbf{I}_a(C)$ , and $D = \mathbf{I}_a(D)$ are collinear. $\blacksquare$

Finally, since $C$ , $L$ , $P$ are collinear, we get $A$ is concyclic with $K = \mathbf{I}_a(C)$ , $L = \mathbf{I}_a(L)$ , $B = \mathbf{I}_a(B)$ , i.e.\ that $AKLB$ is cyclic. So $\overline{KL} \parallel \overline{CD}$ by Reim's theorem, and hence $\overline{PE}$ bisects $\overline{CD}$ by Ceva's theorem.

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khina

993 posts

khina

#15 h Apr 17, 2019, 11:35 PM • 2 Y

Y by megarnie, Adventure10

solution sketch

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shendrew7

792 posts

shendrew7

#102 h Jan 15, 2024, 8:32 PM • 1 Y

Y by L13832

Our solution goes in the reverse direction through phantom points. Suppose $M$ is the midpoint of $CD$ and $Q = ME \cap AB$ . Noting that our desired point $P$ is unqiue by monotonicity, it suffices to show $\angle AQD = \angle BQC$ .

This alternative formulation tells us $\text{[math]}$ is a symmedian of $\triangle EAB$ , so we proceed with length chase:
$\frac{AQ}{QB} = \left(\frac{AE}{EB}\right)^2 = \left(\frac{DE}{EC}\right)^2 = \left(\frac{AD}{BC}\right)^2 = \left(\frac{AB}{BC}\right)^2-1$ $\implies \frac{AB}{QB} = \left(\frac{AB}{BC}\right)^2 \implies BC^2 = BA \cdot BQ.$
Thus we have the angle equality $\angle BQC = \angle BCA$ , and analogously $\angle AQD = \angle BDA$ , so we're done. $\blacksquare$

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OronSH

1727 posts

OronSH

#103 h Jan 30, 2024, 4:02 AM

Y by

had to use a hint buh . The length condition implies that if $Q$ is on $AB$ such that $AD^2=AB\cdot AQ$ then $BC^2=AB\cdot BQ$ so $AD$ is tangent to $(BDQ)$ and $BC$ is tangent to $(ACQ).$ Then angle chasing gives $\angle AQD=\angle ADB=\angle ACB=\angle PQC$ so $P=Q.$ Thus if $PD$ intersects the circumcircle again at $R$ and $BD\cap CP=X$ we see $\angle RBA=\angle RDA=\angle PBD=\angle XBA,$ and $\angle RPB=\angle APD=\angle BPC,$ so $X,R$ are reflections over $AB.$ Similarly if $PC$ intersects the circumcircle again at $S$ and $AC\cap PD=Y$ then $Y,S$ are reflections over $AB.$ Thus $AYXB$ is cyclic and by Reim's $XY\parallel CD$ and Ceva's finishes.

This post has been edited 1 time. Last edited by OronSH, Jan 30, 2024, 1:36 PM

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Bluesoul

872 posts

Bluesoul

#104 h Feb 5, 2024, 6:29 PM

Y by

If $\angle{ADB}=\angle{APD}$ , similar triangle would yield $\frac{AD}{AB}=\frac{AP}{AD}, AD^2=AP\cdot AB; \frac{BC}{BA}=\frac{BP}{BC}, BC^2=AB\cdot BP$ , sum the two equations to get $AD^2+BC^2=AB^2\implies \angle{ADB}=\angle{APD}=\angle{BPC}$ . However, if we vary point $P$ from the point satisfying $\angle{APD}=\angle{CPB}$ , one angle would be larger and the other one would be smaller. Thus, $P$ is unique and we have to prove $PE$ bisects $CD$ .

Since $\angle{ADB}=\angle{ACB}=\angle{CPB}=\angle{APD}$ , it imples $ADSP, BPRC$ are cyclic( $S=BD\cap CP; R=AC\cap DP$ ).

Note $\angle{CRB}=\angle{CPB}=\angle{APD}=\angle{ASD}=180-\angle{ASB}=180-\angle{ARB}\implies \angle{ARB}=\angle{ASB}\implies ARSB$ is cyclic. Finally, note $\angle{BAS}=\angle{BRS}=\angle{SDP}; \angle{BRP}=\angle{BCP}=\angle{BAC}=\angle{BDC}\implies RS||CD$ . Thus $M$ is the midpoint of $CD$ by Ceva's theorem.

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dolphinday

1318 posts

dolphinday

#105 h Feb 12, 2024, 3:19 PM • 1 Y

Y by L13832

Let $P'$ be the point on side $AB$ so that $AP' \cdot AB = AD^2$ , and by PoP we have $AD$ tangent to $(BPP')$ . We have $AD^2 = AB^2 - BC^2 \implies AP' \cdot AB = AB^2 - BC^2 \implies BC^2 = AB \cdot BP'$ so $AC$ is also tangent to $(BP'C)$ . From the length conditions, we have $\triangle AP'D \sim \triangle ABD$ and $\triangle BP'C \sim \triangle BCA$ . From this we get that $\angle BP'C = \angle BCA = \angle BDA = \angle APD$ , so $P' = P$ . Note that $AP \cdot AB = AD^2$ and $AB \cdot BP = BC^2$ implies that $\frac{AP}{BP} = \frac{AD^2}{BC^2} = \frac{AE^2}{BE^2}$ so $PE$ is the $E$ -symmedian of $\triangle BEA$ . This implies that $EM$ is a median of $\triangle CED$ , as desired.

This post has been edited 1 time. Last edited by dolphinday, Mar 17, 2025, 3:40 PM

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pi_is_3.14

1437 posts

pi_is_3.14

#106 h Mar 8, 2024, 6:38 PM

Y by

We prove the converse, something possible by phantom points.

By Ratio Lemma on $\triangle ECD$ with the midpoint (say) M, we have $ED \sin(\angle PEB) = EC \sin(\angle PEA)$ . Now, Ratio Lemma on $\triangle EAB$ gives $\frac{AP}{PB} = \frac{EA \sin(\angle PEA)}{EB \sin(\angle PEB)} = \frac{EA \cdot ED}{EB \cdot EC} = \frac{EA^2}{EB^2} = \frac{AD^2}{BC^2}$ with the last following by $\triangle EAD \sim \triangle EBC$ .

Let $\frac{AD^2}{AP} = \frac{BC^2}{BP} = x$ . Now, $x(AP + BP) = AB^2 \implies x = AB$ or $AD^2 = AP \cdot AB$ and $BC^2 = BP \cdot AB$ . This implies tangents.

Now, $\angle DPA = \angle ADB = \angle ACB = \angle CPB$ .

Motivation: P is an unknown point so it makes sense to try and 'solve for its location' (i.e. find the ratio $\frac{AP}{PB}$ ). The use of the converse is more natural as it allows one to easily find the ratio. No similar triangles or any other length condition can be found directly by the angle condition so using the midpoint is more natural. This motivates the use of trig and allows for a relatively straightforward solution via length bashing.

This post has been edited 4 times. Last edited by pi_is_3.14, Mar 8, 2024, 9:37 PM

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r00tsOfUnity

695 posts

r00tsOfUnity

#107 h Mar 11, 2024, 5:57 PM • 1 Y

Y by L13832

Let $M$ be the midpoint of $\overline{CD}$ , and suppose $\overline{ME}$ intersects $\overline{AB}$ at $X$ . We are left to prove that $X=P$ , i.e., $\angle AXD=\angle BXC$ . Note that $\triangle EAB\stackrel{-}{\sim}\triangle EDC$ and $\overline{EM}$ is the $E$ -median of $\triangle EDC$ , so $\overline{EX}$ is the $E$ -symmedian of $\triangle EAB$ . Thus $\frac{AX}{BX}=\left(\frac{AE}{BE}\right)^{2}=\left(\frac{AD\cdot\tfrac{\sin\angle ADE}{\sin\angle AED}}{BC\cdot\tfrac{\sin\angle BCE}{\sin\angle BEC}}\right)^{2}$ but, as $\angle ADE=\angle BCE$ (cyclic quadrilateral) and $\angle AED=\angle BEC$ (vertical angles), this is just $\left(\tfrac{AD}{BC}\right)^{2}$ . It follows $AX=\tfrac{AD^{2}}{AB}$ and $BX=\tfrac{BC^{2}}{AB}$ ; rearranging gives that $\overline{AD}$ and $\overline{BC}$ are tangent to $(BXD)$ and $(AXC)$ , respectively. Then $\angle AXD=\pi-\angle ADX-\angle DAX=\pi-\angle DBX-\angle DAX=\angle ADB,\quad\text{and}\quad\angle BXC=\pi-\angle BCX-\angle CBX=\pi-\angle CAX-\angle CBX=\angle ACB$ but obviously, $\angle ADB=\angle ACB$ because $ABCD$ is cyclic, $\angle AXD=\angle BXC$ and $X=P$ as desired.

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L13832

250 posts

L13832

#108 h Aug 18, 2024, 8:03 AM • 1 Y

Y by CRT_07

50th post!!! :-D

Let $M$ be the midpoint of $CD$ and let $ME\cap AB=P'$ . So we need to prove $\angle AP'D=\angle CP'B$
$\textbf{Claim 1.}$ $\triangle EDC \sim \triangle EAB$
$\begin{align*} \frac{\sin{\angle P'EA}}{\sin P'EB}=\frac{\sin{\angle MEC}}{\sin{\angle MED}}=\frac{ED}{EC}=\frac{EA}{EB} \end{align*}$ $\therefore EP'$ is the symmedian of $\triangle ABE$ .

Note that
$\begin{align} \frac{AP'}{P'B}&=\frac{EA\cdot \sin{\angle AEP'}}{EB\cdot \sin{\angle BEP'}}=\frac{AE^2}{BE^2}=\frac{DE^2}{CE^2}\\ \frac{AE}{BE}&=\frac{\sin{\angle ABE}}{\sin{\angle BAE}}=\frac{\sin{\angle ABD}}{\sin{\angle BAC}}=\frac{AD}{BC} \end{align}$ From (1) and (2) we have
$\boxed{\frac{AE^2}{BE^2}=\frac{AD^2}{BC^2}=\frac{DE^2}{CE^2}=\frac{AP'}{P'B}}$ So, $\triangle EAD \sim \triangle EBC$ .

$\textbf{Claim 2.}$
$\triangle BCA \sim \triangle BP'C$
Using the fact that $AP'+P'B=AB$ we get:
$\frac{AB}{BP'}=\frac{AB^2}{BC^2}\implies \frac{AB}{BC}=\frac{BC}{BP'}$ Similarly we also get $\triangle BDA\sim \triangle DPA'$ .

By $\textbf{Claim 2.}$ we have
$\angle AP'D=\angle ADB=\angle ACB=\angle CP'B$ $\blacksquare$

This post has been edited 1 time. Last edited by L13832, Aug 18, 2024, 8:05 AM

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Saucepan_man02

1299 posts

Saucepan_man02

#109 h Dec 2, 2024, 2:56 PM

Y by

Here we go:

Redefine $P$ to be a point on $AB$ such that: $AP \cdot AB = AD^2, BP \cdot BA = BC^2$ (which exists uniquely due to the given condition).
It suffice to show that: $\angle APD = \angle BPC$ and $PE$ to be the symmedian of $\triangle EAB$ .
Claim: $\angle APD = \angle BPC$
Note that: $\triangle APD \sim \triangle ADB, \triangle BPC \sim BCA$ . Thus: $\angle APD = \angle ADB = \angle ACB = \angle BPC.$
Claim: $PE$ is a symmedian of $\triangle EAB$ .
Note that; $AP \cdot AB = AD^2, BP \cdot BA = BC^2$ . Dividing them gives us: $\frac{AP}{PB} = \frac{AD^2}{BC^2} = \frac{AE^2}{BE^2}$ and therefore we are done.

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blueprimes

306 posts

blueprimes

#110 h Jan 6, 2025, 2:07 PM

Y by

Let $M$ be the midpoint of $\overline{CD}$ , we define the phantom point $P' = ME \cap \overline{AB}$ . To prove $P \equiv P'$ it suffices to show $\angle AP'D = \angle BP'C$ . Suppose $N$ is the midpoint of $\overline{AB}$ , clearly $\angle AEN = \angle DEM = \angle BEP'$ so $N$ and $P'$ are isogonal conjugates w.r.t. $E$ in $\triangle AEB$ , then $EP'$ is the $E$ -symmedian. By a well-known fact,
$\dfrac{AP'}{BP'} = \dfrac{AE^2}{BE^2} = \dfrac{AD^2}{BC^2} \implies \dfrac{AP' + BP'}{BP'} = \dfrac{AD^2 + BC^2}{BC^2}$ $\implies \dfrac{AB}{BP'} = \dfrac{AB^2}{BC^2} \implies \dfrac{BP'}{BC} = \dfrac{BC}{AB}$ but $\angle P'BC = \angle CBA$ so by SAS Similarity we have $\triangle BP'C \sim \triangle BCA$ . Analogously, $\triangle AP'D \sim \triangle ADB$ , then
$\angle AP'D = \angle ADB = \angle ACB = \angle BP'C$ as wanted.

This post has been edited 1 time. Last edited by blueprimes, Jan 12, 2025, 12:43 PM

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Ilikeminecraft

302 posts

Ilikeminecraft

#111 h Feb 19, 2025, 12:24 AM

Y by

Draw the circle with center $A$ and radius $AD,$ center $B$ and radius $BC.$ We refer to these as $(A)$ and $(B)$ from here. By the given condition, these two circles are orthogonal.
Claim: $P$ lies on the common chord/radax of $(A), (B).$
Proof: Let the common chord intersect $AB$ at $P'.$ Clearly, inverting about $(B)$ results in $A\leftrightarrow P',$ and so $\angle BP'C = \angle BCA = \angle BDA.$ Inverting about $(A)$ results in $\angle BDA = \angle DP'A.$ Hence, $P = P'.$

Let $K$ be the inverse of $C$ w.r.t. $(A).$ Obviously, $K$ lies on $(B),$ but we also have that $\angle KPA = \angle ACB$ from inversion, and thus, $\angle KPA = \angle ACB = \angle BDA = \angle DPA,$ so $D, K, P$ are collinear. Let $L$ be the inverse of $B$ w.r.t. $(B).$ We can also deduce $P, L, C$ are collinear.

Now, note that $\angle ALB = \angle ALP + \angle PLB = \angle ADP + \angle DAP = \angle DPB,$ where we use the fact that $DLPA$ is cyclic. We also have that $\angle CPA = \angle AKB,$ so $AKLB$ is cyclic. Hence, $\angle DLK = \angle CAB = \angle CDL,$ so $CD\parallel KL.$

Now, let $M = PE \cap CD.$ We have that $KP\cdot LC \cdot MD = LP \cdot MC \cdot DK$ from Ceva's, but by parallel, we get $MD = MC$ .

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cj13609517288

1869 posts

cj13609517288

#112 h Feb 20, 2025, 9:57 PM

Y by

Let $Q$ be the point on $AB$ such that $AQ\cdot AB=AD^2$ , and therefore $BQ\cdot BA=BC^2$ . Then
$\angle AQD=\angle ADB=\angle BCA=\angle BQC.$ But it is not hard to show that $\angle AQD-\angle BQC$ is monotonic when moving along $AB$ , so $P=Q$ .

Now define $S=AC\cap DP$ , $T=BD\cap CP$ , and $C',D'$ as the reflection of $C,D$ over $AB$ , respectively. Then
$\angle ACD=\angle ABD=\angle ADP=\angle ADS\Longrightarrow \triangle ADS\sim\triangle ACD,$ so
$\angle AD'C'=\angle ADC=\angle ASD.$ So $S$ lies on $(ABC'D')$ , and similarly $T$ does as well. Now
$\angle ATD'=\angle ABD'=\angle ABD=\angle ABT,$ so $AD=AD'=AT$ , and similarly $BC=BC'=BS$ .

At this point, we have all of the necessary ingredients to cook up a length chase. We have
$\frac{TP}{TB}=\frac{AT}{AB}=\frac{AP}{AT},$ $\frac{SP}{SA}=\frac{BS}{BA}=\frac{BP}{BS}.$ But note that
$\left.\frac{TP}{TB}\middle/\frac{SP}{SA}\right.= \left.\frac{\sin\angle ABT}{\sin\angle BPT}\middle/\frac{\sin\angle BAS}{\sin\angle APS}\right.=\frac{\sin\angle ABT}{\sin\angle BAS}=\frac{AE}{BE}.$ So we have
$\frac{AT}{BS}=\frac{AE}{BE},$ so
$\frac{AE}{BE}=\left.\frac{AP}{AT}\middle/\frac{BP}{BS}\right.=\frac{AP}{BP}\cdot\frac{BE}{AE}\Longrightarrow\frac{AP}{BP}=\left(\frac{AE}{BE}\right)^2.$ So $EP$ is the $E$ -symmedian of triangle $EAB$ , thus the $E$ -median of triangle $ECD$ . $\blacksquare$

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pi271828

3363 posts

pi271828

#113 h Feb 23, 2025, 3:28 AM

Y by

Claim: There is a unique point $P$ on side $\overline{AB}$ that satisfies $\angle APD = \angle BPC$ .

Proof. Vary $P$ across side $\overline{AB}$ from $A$ to $B$ . Note that $\angle APD$ is decreasing, while $\angle BPC$ is increasing, so $P$ is unique. $\square$

Claim: Let $Q$ be the foot of the $E$ -symmedian in $\triangle AEB$ . We claim that $Q$ is the unique point satisfying the angle condition.

Proof. Note that $\begin{align*} \frac{QA}{QB} = \frac{EA^2}{EB^2} = \frac{AD^2}{BC^2} = \frac{AB^2}{BC^2} - 1\end{align*}$ so we must have $\begin{align*} \frac{AB}{QB} = \frac{AB^2}{BC^2}\end{align*}$ which implies $BC^2 = PB \cdot AB$ , and therefore $BC$ is tangent to $(APC)$ . Similarly, $AD$ is tangent to $(BPD)$ . From these tangencies, we have $\begin{align*} \angle APD = \angle ADB = \angle ACB = \angle CPB \end{align*}$ so the claim the proven. $\square$

To finish, note that the $E$ -symmedian of $\triangle AEB$ is the $E$ -median of $\triangle ECD$ because $AB$ and $CD$ are anti-parallel, so we are done. $\blacksquare$

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YaoAOPS

1496 posts

YaoAOPS

#114 h Mar 10, 2025, 11:16 PM

Y by

Since $AD^2 + BC^2 = AB^2$ it follows that for some point $P'$ , $(DP'B)$ is tangent to $AD$ and $(CP'A)$ is tangent to $BC$ . Then this implies that
$\measuredangle APD = \measuredangle DPB = \measuredangle ADB = \measuredangle ACB = \measuredangle APC = \measuredangle CPB$ so $P' = P$ . Now let $CD$ intersect $(APC)$ and $(DPB)$ again at $F$ and $G$ . We have that
$CF = CA * \frac{\sin FAC}{\sin AFC} = CA \cdot \frac{\sin DCB}{\sin CPB} = BD \cdot \frac{\sin ADC}{\sin DPA} = DG$ by symmetry so $M$ lies on the radical axis of $(APB)$ and $(CBP)$ , so since $E$ and $P$ do as well the result follows.

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peace09

5416 posts

peace09

#115 h Mar 17, 2025, 8:38 PM

Y by

Pick $P'\in AB$ with $(AP',BP')=(\tfrac{AD^2}{AB},\tfrac{BC^2}{AB})$ ; by SAS $ADP\sim ABD$ and $BCP\sim BAC$ , giving
$\angle APD=\angle ADB=\angle ACB=\angle BPC.$ But there exists a unique $X\in AB$ with $\angle AXD=\angle BXC$ , since the function $f(X):=\angle AXD-\angle BXC$ is monotonic as $X$ varies linearly along $AB$ . So $P'\equiv P$ .

Pick $P''\in AB$ with $EP''$ a symmedian of $\triangle ABE$ ; famously $\tfrac{AP''}{BP''}=(\tfrac{AE}{BE})^2=(\tfrac{AD}{BC})^2$ . But there exists a unique $X\in AB$ with $(\tfrac{AX}{BX})=(\tfrac{AD}{BC})^2$ , since the function $g(X):=\tfrac{AX}{BX}$ is monotonic as $X$ varies linearly along $AB$ . So $P''\equiv P$ .

Finally, $EP$ is a symmedian of $\triangle ABE$ and therefore a median of $\triangle CDE$ . $\square$

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jcoons91

3 posts

jcoons91

#120 h Mar 17, 2025, 9:28 PM

Y by

My solution:

First, we observe that the point $P$ is uniquely determined. Consider the perpendicular bisector of segment $CD$ and let it intersect line $AB$ at point $X$ . Since the points $C, D, P, X$ are concyclic, it follows that $P$ is the second intersection of the circumcircle of $\odot(CDX)$ and line $AB$ . This establishes the uniqueness of $P$ .

Next, define $M$ as the midpoint of segment $CD$ , and let $P$ be redefined as the intersection of line $EM$ and line $AB$ . To complete the proof, it suffices to show that the angles $\angle APD$ and $\angle BPC$ are equal.

To this end, we observe that the triangles $\triangle AEB$ and $\triangle CED$ are inversely similar, which implies that $P$ is the foot of the symmedian of $\triangle AEB$ . Additionally, given that $\triangle AED \sim \triangle BEC$ , we obtain the proportion:

$\frac{AP}{PB} = \frac{AE^2}{BE^2} = \frac{AD^2}{BC^2}.$
From this length condition, we derive the relationships:

$AP \cdot AB = AD^2, \quad \text{and} \quad BP \cdot BA = BC^2.$
By these equalities, we conclude that the desired angle relation holds:

$\angle APD = \angle ADB = \angle ACB = \angle BPC.$
And we are done. $\square$

This post has been edited 2 times. Last edited by jcoons91, Mar 17, 2025, 9:28 PM
Reason: issues with format

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