Perpendicular diagonals in cyclic quadrilateral

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Perpendicular diagonals in cyclic quadrilateral

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Source: 2014 China TST 1 Day 1 Q1

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61plus

#1 h Mar 18, 2014, 4:36 PM • 5 Y

Y by narutomath96, aopser123, Davi-8191, Adventure10, Rounak_iitr

$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$ , the parallel line $EF$ to $AC$ intersect $AB$ at point $E$ , line $FG$ parallel to $BD$ intersect $CD$ at $G$ . Let the projection of $E$ onto $CD$ be $P$ , projection of $F$ onto $DA$ be $Q$ , projection of $G$ onto $AB$ be $R$ . Prove that $QF$ bisects $\angle PQR$ .

This post has been edited 1 time. Last edited by 61plus, Mar 30, 2014, 5:03 AM

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#2 h Mar 18, 2014, 5:48 PM • 7 Y

Y by narutomath96, R8450932, mijail, Adventure10, Mango247, Rounak_iitr, and 1 other user

Too easy for China.

Lemma 1: Suppose $EP$ meets $BD$ at $K$ . Then $F,K,Q$ are collinear.

Proof: It suffices to show $FK \perp AD$ . Let $AC \cap BD=J$ . We have $\angle KJC= \angle KPC=90^{\circ}$ so $KJPC$ is cyclic and $\angle BKE = \angle JKP=\angle JCP=\angle KBE$ making $\triangle BEK$ $E-$ isosceles and thus making $EF \perp BK$ the perpendicular bisector of $BK$ . Let $EF \cap BK=T$ . Then $\angle KFE=\angle EFB=\angle ACB=\angle ADB$ . If $FK$ hits $AD$ at $Q'$ , we immediately get $\angle Q'DT=\angle Q'FT \implies Q'DTF$ is cyclic and $\angle FTD=90^{\circ}=\angle FQ'D$ meaning $Q=Q'$ and $F,K,Q$ are collinear,

Similarly, if $GR$ cuts $AC$ at $L$ , then $F,L,Q$ are collinear.

Now, $LQAR$ is cyclic, which gives $\angle RQF=\angle BAC$ . Since $KQDP$ is also cyclic, $\angle PQF=\angle CDB$ . But ofcourse $\angle BAC=\angle CDB$ , so $\angle PQF=\angle RQF$ .

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XmL

#3 h Mar 19, 2014, 1:04 AM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

My proof doesn't require proving the collinearity that ThirdTimeLucky mentioned(even though it's pretty apparent when drawn). It's clear that $G,P,F,R,E$ are concyclic, since $\angle BEF=\angle BAC=\angle BDC=\angle CGF$ , therefore $PF=RF$ . This means we only have to prove $Q$ is concyclic with $GPFRE$ . Through $E$ we construct a line parallel to $BD$ and it intersects $AD$ at $X$ , since $\frac {AX}{DX}=\frac {AE}{BE}=\frac {CF}{BF}=\frac {CG}{GD}$ , therefore $GX\parallel AC\Rightarrow FEXG$ is a rectangle. Since $\angle FGX=90$ , hence $Q$ is concyclic with $GPFRE$ and we are done.

This post has been edited 1 time. Last edited by XmL, Mar 25, 2014, 3:54 AM

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elim

#4 h Mar 25, 2014, 3:23 AM • 3 Y

Y by XmL, Adventure10, Mango247

XmL wrote:

My proof doesn't require proving the collinearity that ThirdTimeLucky mentioned(even though it's pretty apparent when drawn). It's clear that $G,P,F,R,E$ are concyclic, since $\angle BEF=\angle BAC=\angle BDC=\angle CGF$ , therefore $PF=CF$ . This means we only have to prove $Q$ is concyclic with $GPFRE$ . Through $E$ we construct a line parallel to $BD$ and it intersects $AD$ at $X$ , since $\frac {AX}{DX}=\frac {AE}{BE}=\frac {CF}{BF}=\frac {CG}{GD}$ , therefore $GX\parallel AC\Rightarrow FEXG$ is a rectangle. Since $\angle FGX=90$ , hence $Q$ is concyclic with $GPFRE$ and we are done.

Don't see $PF=CF$ .

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XmL

#5 h Mar 25, 2014, 3:54 AM • 2 Y

Y by Adventure10, Mango247

elim wrote:

XmL wrote:

My proof doesn't require proving the collinearity that ThirdTimeLucky mentioned(even though it's pretty apparent when drawn). It's clear that $G,P,F,R,E$ are concyclic, since $\angle BEF=\angle BAC=\angle BDC=\angle CGF$ , therefore $PF=CF$ . This means we only have to prove $Q$ is concyclic with $GPFRE$ . Through $E$ we construct a line parallel to $BD$ and it intersects $AD$ at $X$ , since $\frac {AX}{DX}=\frac {AE}{BE}=\frac {CF}{BF}=\frac {CG}{GD}$ , therefore $GX\parallel AC\Rightarrow FEXG$ is a rectangle. Since $\angle FGX=90$ , hence $Q$ is concyclic with $GPFRE$ and we are done.

Don't see $PF=CF$ .

Sorry I meant $PF=RF$

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bojler

#6 h Apr 8, 2014, 4:15 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ and $N$ be midpoints of $AB$ and $BC$ respectively, $S$ be intersection of $AC$ and $BD$ , $M'$ be intersection of $MS$ and $DC$ and $N'$ be intersection of $NS$ and $AD$ .
Angle chase gives us that $MM'$ is perpendicular to $DC$ and $NN'$ perpendicular to $AD$ . Let $X$ be intersection of $EP$ and $FQ$ . Triangles $MNS$ and $EFX$ are similar and they're homothetical with center $B$ (because $EF$ parallel to $MN$ , $FX$ parallel to $NS$ and $EX$ parallel to $MS$ ),so we have $X$ is on $BD$ . Analogously,we have that $Y$ ,intersection of $GR$ and $FQ$ ,lies on $AC$ .
Note that quadrilatelars $AQYR$ and $QXPD$ are cyclic,so we have
$\angle RQF=\angle SAB=90-\angle ABD$ and
$\angle PQF=\angle PDS=90-\angle ABD$ ,and we are done.

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#7 h Apr 24, 2014, 4:29 PM • 2 Y

Y by Adventure10, Mango247

if $FQ \cap BD=X$ notice $FE$ bisects $BX$ and so $EXP$ are collinear. Similarly if $Y=EQ \cap AC$ then $RYG$ are collinear. Notice $RYQA$ and $PXQD$ are cyclic and so $\angle RQY = \angle RAC = \angle XDP = \angle XQP$ and we're done.

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#8 h Jul 27, 2014, 11:44 PM • 2 Y

Y by Adventure10, Mango247

It is obvious that $REFPG$ is cyclic.Now from simple angle chasing we obtain $FP=RF$ .Now denote $X$ be the point of intersection of $BD$ and $EP$ . $XSPC$ is cyclic so we conclude $\triangle EBX$ is isosceles $\implies \triangle FXB$ is isosceles which implies $\angle XFB=\angle ACB$ .Notice that $\angle BAC + \angle BDA + \angle DBA +\angle DBC =180$ and that $\angle EAD=\angle DAC +\angle BAC$ and $\angle AEF=180 - \angle BAC$ .Now we see that $\angle EAQ + \angle AEF + \angle EFX=270$ so $FX$ is perpendicular to $AD$ which implies $X$ lies on $FQ$ .Also let $Y$ be the intersection of $AC$ and $GR$ and then by analogy we have that $Y$ lies on $FQ$ .The problem is now trivial since both $AQYR$ and $DQXP$ are cyclic.

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#9 h Dec 11, 2016, 5:23 PM • 2 Y

Y by Adventure10, Rounak_iitr

Claim: $BD,EP,FQ$ concur.

Proof: Let $EP\cap BD=E',BD\cap EF=L.$ Then $\angle BEF=\angle BAC=\angle BDC=\angle E'DP=\angle E'EL=\angle E'EF$ so it follows that $EL$ is both an altitude and angle bisector of $\triangle BEE',$ and hence $E'$ is the point of $BD$ such that $LE'=BL.$ Similarly, if $FQ\cap BD=F'$ we get a similar conclusion, so $E'=F'$ as desired.

We can also get that $RG,FQ,AC$ concur; let $RG\cap FQ\cap AC=X,BD\cap EP\cap FQ=Y.$ Then $ARXQ, QYPD$ cyclic so $\angle RQX=\angle RAX=\angle RAC=\angle BAC=\angle BDC=\angle YDP=\angle YQP=\angle FQP,$ done.

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#10 h Apr 26, 2019, 2:21 PM • 2 Y

Y by Adventure10, Mango247

Consider the following projective transformations from $AD$ to $AB$ :
1) $h$ is defind as follows:To a point $K$ on $AD$ we find the intersection of the perpendicular to $AD$ at $K$ and $BC$ ,then this point we project parallel to $AC$ and finally make orthogonal projection of this point on $CD$ .
2)We define $g$ as follows:To a point $K$ on $AD$ we find the intersection of the perpendicular to $AD$ at $K$ and $BC$ . Then we project it parallel to $BD$ ,make an orthogonal projection on $AB$ ,say $H$ and finally we find an intersection of the reflection of the line $HK$ across the perpendicular to $AD$ at $K$ and $AB$ .
It’s clear that both transformations preserves cross-ratio.Thus,they are projective and they can be uniquely determined by three pairs of points. So we just need to prove the result for three arbitrary cases.
1. Consider the point $Q$ to be the orthogonal projection of the intersection of diagonals on $AD$ . Then , using Brahmagupta theorem , we conclude that the intersecrion of the perpendicular is the midpoint of $BC$ .So it’s easy to see that if $G$ is the intersection of the diagonals then $AQGR$ and $DQGP$ are cyclic and the conclusion follows.
2.Let $Q$ be the orthogonal projection of $C$ on $AD$ in this case we see that $AQGR$ and $DQGP$ are cyclic( just two opposit angles are $90$ .
3. Let $Q$ be the orthogonal projection of $B$ on $AD$ . This case is analogical to 2.
So we have $h$ = $g$ and we are done.

This post has been edited 1 time. Last edited by NikitosKh, Apr 26, 2019, 2:21 PM

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#11 h Oct 1, 2019, 7:16 AM • 1 Y

Y by Adventure10

Solution

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#12 h Apr 5, 2022, 9:48 PM • 2 Y

Y by centslordm, Rounak_iitr

Claim: $\overline{BD},\overline{PE},$ and $\overline{QF}$ are concurrent.
Proof. Let $X=\overline{PE}\cap\overline{BD}$ and $Q'=\overline{FX}\cap\overline{AD}.$ It suffices to prove $\overline{FQ'}\perp\overline{AD}.$ Notice $\measuredangle FEB=\measuredangle CAB=\measuredangle CDB=90-\measuredangle DXP=\measuredangle XEF$ so $\triangle EBF\cong\triangle EXF.$ Hence, $\measuredangle Q'XD+\measuredangle XDQ'=\measuredangle FXB+\measuredangle BCA=\measuredangle XBF+\measuredangle BFE=90.$ $\blacksquare$

Similarly, let $Y=\overline{FQ}\cap\overline{GR}\cap\overline{AC}.$ Then, $DQXP$ and $ARYQ$ are cyclic, so $\measuredangle PQX=\measuredangle PDX=\measuredangle CAB=\measuredangle YQR.$ $\square$

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#13 h May 24, 2024, 6:35 AM

Y by

I'm actually tweaking from China geo.

The main claim is that $E$ , $F$ , $G$ , $P$ , $Q$ and $R$ are concyclic. This just follows from,
$\begin{align*} \angle EFG = \angle ERG = \angle EPG = 90 \end{align*}$ and from noting that $Q$ lies on this circle with easy complex bash. Namely $f = kb + (1 - k)c$ , $e = kb + (1 - k)a$ , $g = kd + (1 - k)c$ and then use complex foot to get $q = \frac{1}{2}(a + d + f - ad\overline{f})$ for real $k$ . The remaining bash isn't bad, and I am too lazy too type it up. Then angle chase using parallel lines and done.

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#14 h Sep 29, 2024, 1:15 PM • 2 Y

Y by Rounak_iitr, bo18

Let C' is the reflection of C in GF. Then C' lies on AC, because CC' || AC. Then let Q' is the intersection of FC' and AD.
$\angle DQ'F = 360 - \angle Q'DC - \angle DCF - \angle DFQ' = 360 - \angle ADB - \angle BDC - \angle DCB - 2\angle CFG = 360 - (\angle CFG + \angle BDC + \angle DCB) - \angle CFG - \angle ADB = 180 - (\angle ADB + \angle CAD) = 90.$
So Q=Q' and C' lies on FQ. Because of the symmetry C' lies on GR. Now let B' is the reflection of B in EF. The B' lies on BD, FQ and EP.
Now we have cyclic ARC'Q and DQB'P.
$\angle PQF = \angle PQB' = \angle PDB' = \angle CDB = \angle CAB = \angle C'AR = \angle C'QR = \angle FQR$

This post has been edited 1 time. Last edited by africanboy, Sep 29, 2024, 1:15 PM

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