413. Geometry 6.

by Virgil Nicula, Mar 21, 2015, 1:43 PM

PP2. Let $\triangle ABC$ and the points $M\in (BC)\ ,\ N\in (MC).$ Suppose that there are $\left\{\begin{array}{ccc} X\in [AB & ; & Y \in [AM\\\\ Z\in [AN & ; & T \in [AC\end{array}\right\|$ so that $XY=YZ=ZT.$ Prove that $3\cdot MN\le BC.$

Proof. Let $\left\{\begin{array}{ccc} m\left(\widehat{XAY}\right) & = & \alpha\\\ m\left(\widehat{YAZ}\right) & = & \beta\\\ m\left(\widehat{ZAT}\right) & = & \gamma\end{array}\right\|.$ Thus $:\ \left\{\begin{array}{ccccccc} \frac {YX}{YZ} & = & \frac {AX}{AZ}\cdot\frac {\sin\widehat{YAX}}{\sin\widehat{YAZ}} & \iff & \frac {YX}{YZ} & = & \frac {AX}{AZ}\cdot\frac {\sin\alpha}{\sin\beta}\\\\ \frac {TX}{TZ} & = & \frac {AX}{AZ}\cdot\frac {\sin\widehat{TAX}}{\sin\widehat{TAZ}} & \iff & \frac {TX}{TZ} & = & \frac {AX}{AZ}\cdot\frac {\sin (\alpha +\beta +\gamma )}{\sin\gamma}\end{array}\right\|$ $\implies$ $\frac {YX}{YZ}:\frac {TX}{TZ}=\frac {\sin\alpha\sin\gamma}{\sin\beta\sin (\alpha +\beta +\gamma )}=k\ (*)$ is constant,

i.e. $(\forall )$ $\left\{\begin{array}{ccc} X\in [AB & ; & Y\in [AM\\\\ Z\in [AN & ; & T\in [AC\end{array}\right\|$ the expression $\frac {YX}{YZ}:\frac {TX}{TZ}$ is constant. Notam $BM=m,$ $MN=n,$ $NC=p.$ Hence, $XY=YZ=ZT=x$ $\implies$ $\frac {YX}{YZ}:\frac {TX}{TZ}=\frac 13$

$\implies k=\frac 13$ $\implies$ $\frac {MB}{MN}:\frac {CB}{CN}=\frac 13$ $\implies$ $\frac mn:\frac {m+n+p}p=\frac 13$ $\implies$ $n(m+n+p)=3mp\implies$ $n^2\cdot \frac{m+n+p}3=mnp\implies$ $n^2\cdot \frac {m+n+p}3\le \left(\frac {m+n+p}3\right)^3\implies$

$n^2\le \left(\frac {m+n+p}3\right)^2\implies$ $n\le \frac {m+n+p}3\implies$ $3\cdot MN\le BC.$

Extension. Let $\triangle ABC$ and $M\in (BC),$ $N\in (MC).$ Suppose that there are $\left\{\begin{array}{ccc} X\in [AB & ; & Y \in [AM\\\\ Z\in [AN & ; & T \in [AC\end{array}\right\|$ so that $2\cdot YZ=XY+ZT.$ Prove that $2\cdot MN\le BM+NC.$

PP3 (O.I.M. - 2012). Let $\triangle ABC$ with $A=90^{\circ}$ and $AD\perp BC$ , where $D\in (BC)$ . For $X\in (AD)$ define

$\left\{\begin{array}{ccc} K\in (CX)\ ; & BK=BA\\\\ L\in (BX)\ ; & CL=CA\end{array}\right\|$ and $M\in CL\cap BK$ . Show that $MK=ML$ (Josef Tkadlec, Czech).

Proof (metrical). Denote $Y\in XM\cap BC$ and $AD=h$ , $XD=p\in (0,h)$ where $ah=bc$ . Thus, $\overline{AXD}\perp BC\implies$ $BA^2-BX^2=CA^2-CX^2=$ $DA^2-DX^2=$

$h^2-p^2\implies$ $\left\{\begin{array}{ccc} BX^2 & = & c^2+p^2-h^2\\\\ CX^2 & = & b^2+p^2-h^2\end{array}\right\|$ . Denote $\left\{\begin{array}{ccc} \frac {LB}m=\frac {LX}1=\frac {BX}{m+1}\\\\ \frac {KC}n=\frac {KX}1=\frac {CX}{n+1}\end{array}\right\|$ and apply the Stewart's relation to the the cevians $CL$ , $BK$ in the triangle $BXC\ :$

$\blacktriangleright\ CL^2\cdot BX+BX\cdot LB\cdot LX=$ $CX^2\cdot LB+CB^2\cdot LX$ $\implies$ $b^2+\frac m{(m+1)^2}\cdot BX^2=$ $CX^2\cdot\frac m{m+1}+a^2\cdot \frac 1{m+1}\implies$

$b^2(m+1)^2+m\left(c^2+p^2-h^2\right)=$ $m(m+1)\left(b^2+p^2-h^2\right)+a^2(m+1)\implies$ $m^2\left(h^2-p^2\right)+\left(b^2-a^2\right)=0\implies$ $\boxed{\frac {LB}{LX}=m=\frac c{\sqrt{h^2-p^2}}}\ (1)\ .$

$\blacktriangleright\ BK^2\cdot CX+CX\cdot KC\cdot KX=$ $BX^2\cdot KC+BC^2\cdot KX\implies$ $c^2+\frac n{(n+1)^2}\cdot CX^2=$ $BX^2\cdot\frac n{n+1}+a^2\cdot \frac 1{n+1}\implies$

$c^2(n+1)^2+n\left(b^2+p^2-h^2\right)=$ $n(n+1)\left(c^2+p^2-h^2\right)+a^2(n+1)\implies$ $n^2\left(h^2-p^2\right)+\left(c^2-a^2\right)=0\implies$ $\boxed{\frac {KC}{KX}=n=\frac b{\sqrt{h^2-p^2}}}\ (2)\ .$

Hence $(1)\ \wedge\ (2)\ \implies\ \boxed{\frac bn=\frac cm=\sqrt{h^2-p^2}}\ (*)$ . Apply the Ceva's theorem to $M$ and $\triangle BXC\ :\ \frac {YB}{YC}\cdot\frac {KC}{KX}\cdot\frac {LX}{LB}=1\iff$ $\frac {YB}{YC}\cdot \frac b{\sqrt{h^2-p^2}}\cdot\frac {\sqrt{h^2-p^2}}c=1\iff$

$\boxed{\frac {YB}{YC}=\frac cb}\ \implies\ Y$ is the foot of the $A$ -bisector in $\triangle ABC$ , i.e. $(\forall )\ p\in (0,h)$ the line $XM$ pass by a fixed point $Y$ . Apply the Menelaus' theorem to the mentioned transversals $:$

$\left\{\begin{array}{ccccc} \blacktriangleright\ \overline {BLX}/\triangle CMY\ : \frac{BY}{BC}\cdot\frac {LC}{LM}\cdot \frac {XM}{XY}=1 & \implies & \frac c{b+c}\cdot \frac b{ML}\cdot \frac {XM}{XY}=1 & \implies & \boxed{ML=\frac {bc}{b+c}\cdot \frac {XM}{XY}}\\\\ \blacktriangleright\ \overline {CKX}/\triangle BMY\ : \frac{CY}{CB}\cdot\frac {KB}{KM}\cdot \frac {XM}{XY}=1 & \implies & \frac b{b+c}\cdot \frac c{KM}\cdot \frac {XM}{XY}=1 & \implies & \boxed{MK=\frac {bc}{b+c}\cdot \frac {XM}{XY}}\end{array}\right\|$ $\implies$ $\boxed{MK=ML=\frac {bc}{b+c}\cdot \frac {XM}{XY}}$ .

Remark. $m\left(\widehat{BXC}\right)=\phi\implies$ $4[BXC]=\left(XB^2+XC^2-BC^2\right)\tan\phi\implies$ $2pa=\left[\left(c^2+p^2-h^2\right)+\left(b^2+p^2-h^2\right)-a^2\right]\tan\phi$ $\implies$ $\boxed{\tan\phi =\frac {pa}{p^2-h^2}}$ .

An easy extension. Let an acute $\triangle ABC$ with $D\in BC\ ,\ AD\perp BC$ and $X\in (AD)$ . Define $\left\{\begin{array}{c} K\in (CX)\ ,\ BK=BA\\\\ L\in (BX)\ ,\ CL=CA\end{array}\right\|$

and $\left\{\begin{array}{c} M\in CL\cap BK\\\\ Y\in XM\cap BC\end{array}\right\|$ . Show that $\frac {YB}{YC}=\frac {\sqrt{a^2-b^2}}{\sqrt{a^2-c^2}}$ and $\frac {MK}{b\sqrt{a^2-b^2}}=\frac {ML}{c\sqrt{a^2-c^2}}=\left(\sqrt{a^2-b^2}+\sqrt{a^2-c^2}\right)\cdot\frac {XM}{XY}$ .

PP4. Let an acute $\triangle ABC$ with the orthocenter $H$ and $\left\{\begin{array}{ccc} E\in AC\ ; & BE\perp AC\\\\ F\in AB\ ; & CF\perp AB\\\\ X\in (BF)\ ; & Y\in (CE)\end{array}\right\|$ so that $H\in (XY)$ denote $\left\{\begin{array}{cc} U\in (BH)\ ; & XU\perp XY\\\\ V\in (CH)\ ; & YV\perp XY\end{array}\right\|$ . Prove that $UV\parallel BC$ .

Proof. Observe that $BCEF$ , $UXEY$ and $VXFY$ are cyclically $\implies$ $\left\{\begin{array}{ccc} \widehat{EFC} & \equiv & \widehat{EBC}\ (*)\\\\ HU\cdot HE & = & HX\cdot HY\\\\ HV\cdot HF & = & HX\cdot HY\end{array}\right\|\implies$

$HU\cdot HE=HV\cdot HF\implies$ $UVEF$ is cyclically $\implies$ $\widehat{EFC}\equiv \widehat{EUV}\ \stackrel{(*)}{\implies}\ \widehat{EBC}\equiv$ $\widehat{EUV}\implies UV\parallel BC$ .

PP5. Let the circles $u=\mathbb C(I)$ , $v=\mathbb C(J)$ with $\{A,B\}\subset u\cap v$ and $M\in (AB)$ , $X\in u$ , $Y\in v$ so that $M\in (XY)$ and $MX=MY$ . Prove that $XI^2+XJ^2=YI^2+YJ^2$ .

Proof. Denote $\left\{\begin{array}{ccc} \{X,V\} & = & \{X,M\}\cap u\\\\ \{Y,U\} & = & \{Y,M\}\cap v\end{array}\right\|$ and the power $p_w(X)$ of the point $X$ w.r.t. the circle $w$ . Thus, $AB$ is the radical axis of $u$ , $v$ and $M\in (AB)$ $\implies$

$p_u(M)=p_v(M)$ $\implies$ $MV\cdot MX=MU\cdot MY$ $\implies$ $MU=MV$ $\implies$ $MX-XU=MY-YV$ $\implies$ $XU=YV$ $\implies$ $XY\cdot XU=YX\cdot YV$ $\implies$

$p_v(X)=p_u(Y)$ $\implies$ $XJ^2-YJ^2=YI^2-XI^2$ $\implies$ $XI^2+XJ^2=YI^2+YJ^2$ .

PP6. Let $A$ , $B$ and $O$ be three points so that $OA=a$ and $OB=b$ . Let a mobile point $P$ which belongs to the circle $w=\mathbb C(O,r)$ . Find the maximum value of the sum $PA+PB$ .

Proof. Let $PA=x$ , $PB=y$ . Apply the Stewart's to $PO$ in $\triangle APB\ :\ PA^2\cdot OB+PB^2\cdot OA=PO^2\cdot AB+AB\cdot OA\cdot OB\iff$ $\boxed{bx^2+ay^2=(a+b)\left(r^2+ab\right)}\ (*)$

(constant). Denote $\boxed{x+y=\lambda>0}$ . Obtain the equation $bx^2+a(\lambda -x)^2-(a+b)\left(r^2+ab\right)=0$ , i.e. $\boxed{(a+b)x^2-2a\lambda x+a\lambda ^2-(a+b)\left(r^2+ab\right)=0}$ . This equation

has real roots $\iff$ $\Delta^{\prime}=a^2\lambda^2-a(a+b)\lambda^2+(a+b)^2\left(r^2+ab\right)\ge 0\iff$ $(a+b)^2\left(r^2+ab\right)\ge ab\lambda^2$ , i.e. $\boxed{\lambda\le (a+b)\sqrt{\frac {r^2}{ab}+1}}$ . Therefore, $\lambda_{\mathrm{max}}=(a+b)\sqrt{\frac {r^2}{ab}+1}$

and $\frac {x_{\mathrm{max}}}a=\frac {y_{\mathrm{\max}}}b=\frac {\lambda_{\mathrm{max}}}{a+b}=\sqrt{\frac {r^2}{ab}+1}$ . In conclusion, the maximum value of the sum $PA+PB$ is equally to $(a+b)\sqrt{\frac {r^2}{ab}+1}$ .

PP7 (Norvegia - 2014). Let $ABCD$ be a parallelogram and $\left\{\begin{array}{ccc} M\in (BC)\\\\ N\in (CD)\end{array}\right\|$ so that $BM=DN$ and $P\in BN\cap DM$ . Prove that the ray $[AP$ is the bisector of $\widehat{BAD}$ .

Proof 1. $\left\{\begin{array}{c} R\in BP\cap AD\\\\ X\in AP\cap BC\end{array}\right\|$ and $\left\{\begin{array}{c} AD=BC=a\\\\ AB=CD=b\\\\ BM=DN=m\end{array}\right\|\implies$ $RD=\frac {ma}{b-m}$ and $\widehat{BXA}\equiv\widehat{XAD}$ . Thus, $\frac {MX}{AD}=\frac {PM}{PD}=\frac{BM}{DR}\implies$ $\frac {MX}a=\frac m{\frac {ma}{b-m}}\implies$

$MX=b-m\implies$ $BX=BM+MX\implies BX=b\implies BX=BA\implies$ $\widehat{BXA}\equiv\widehat{BAX}\implies$ $\widehat{BAX}\equiv \widehat{XAD}\implies$ $[AP$ is the bisector of the angle $\widehat{BAD}$

Proof 2. $\left\{\begin{array}{c} S\in AP\cap BD\\\\ R\in BP\cap AD\\\\ Q\in DM\cap AB\end{array}\right\|$ and $\left\{\begin{array}{c} AD=BC=a\\\\ AB=CD=b\\\\ BM=DN=m\end{array}\right\|\implies$ $\left\{\begin{array}{ccccccccc} BM\parallel AD & \iff & \frac {BM}{AD}=\frac {QB}{QA} & \iff & \frac ma=\frac {QB}{QB+b} & \iff & \frac m{a-m}=\frac {QB}b & \iff & QB=\frac {mb}{a-m}\\\\ DN\parallel AB & \iff & \frac {DN}{AB}=\frac {RD}{RA} & \iff & \frac mb=\frac {RD}{RD+a} & \iff & \frac m{b-m}=\frac {RD}a & \iff & RD=\frac {ma}{b-m}\end{array}\right\|$ .

Observe that $\left\{\begin{array}{ccc} AR=AD+RD=a+\frac {ma}{b-m} & \implies & AR=\frac {ab}{b-m}\\\\ AQ=AB+QB=b+\frac {mb}{a-m} & \implies & AQ=\frac {ab}{-m}\end{array}\right\|$ . Apply the Menelaus' theorem to the transversal $\overline{ASP}/\triangle BDR\ :\ \frac {AD}{AR}\cdot\frac {PR}{PB}\cdot\frac {SB}{SD}=1$

$\iff$ $\frac {AD}{AR}\cdot\frac {DR}{BM}\cdot\frac {SB}{SD}=1$ $\iff$ $\frac {b-m}b\cdot\frac a{b-m}\cdot\frac {SB}{SD}=1$ $\iff$ $\frac {SB}{SD}=\frac ba$ $\iff$ $\frac {SB}{SD}=\frac {AB}{AD}$ $\iff$ the ray $[AP$ is the bisector of the angle $\widehat{BAD}$ .

Remark. If denote $L\in BD\cap RQ$ , then $AL\perp AP$ . Indeed, the division $(D,S,B,L)$ is harmonically, i.e. $\frac {SB}{SD}=\frac {LB}{LD}$ and in this case $\widehat{SAD}\equiv\widehat{SAB}\implies$ $AP\perp AL$ .

PP8. Let $\triangle ABC$ and the midpoint $M$ of the side $[BC]$ . Denote $\{P,R\}\subset (AM)$ so that $\left\{\begin{array}{ccc} \widehat{PBA} & \equiv & \widehat{PBC}\\\\ \widehat{RCA} & \equiv & \widehat{RCB}\end{array}\right\|$ . Prove that $BP=CR\iff AB=AC$ .

Proof 1. Denote $BC=2m$ . Thus, $BP=CR\iff$ $\frac {2mc\cos\frac B2}{m+c}=\frac {2mb\cos\frac C2}{m+b}\iff$ $m\left(b\cos\frac C2-c\cos\frac B2\right)=$ $bc\left(\cos\frac B2-\cos\frac C2\right)\iff$ $m\left(\frac {\cos\frac C2}c-\frac {\cos\frac B2}b\right)=$

$\cos\frac B2-\cos\frac C2\iff$ $\frac m{4R}\left(\frac 1{\sin\frac C2}-\frac 1{\sin\frac B2}\right)=\cos\frac B2-\cos\frac C2\iff$ $\frac m{4R\sin\frac B2\sin\frac C2}\left(\sin\frac B2-\sin\frac C2\right)=\cos\frac B2-\cos\frac C2$ . Define for $(\forall )\ \{x,y\}\subset\mathbb R$ the equivalence

relation $"\underline{\mathrm{same\ sign}}"\ : x\ .s.s.\ y\ \iff\ xy>0\ \vee\ x=y=0$ . Therefore, $\left(\frac B2-\frac C2\right)$ $\ .s.s.\ \left(\sin\frac B2-\sin\frac C2\right)$ $\ .s.s.\ \left(\cos\frac B2-\cos\frac C2\right)$ $\ .s.s.\ \left(\frac C2-\frac B2\right)\implies$

$\left(\frac B2-\frac C2\right)\ .s.s.\ \left(\frac C2-\frac B2\right)\iff$ $\left(\frac B2-\frac C2\right)\left(\frac C2-\frac B2\right)\ge 0\iff$ $(B-C)(C-B)\ge 0\iff$ $(B-C)^2\le 0\iff B=C$ .

Proof 2. Denote $I\in BP\cap CR$ and $\left\{\begin{array}{cc} m\left(\widehat{BPM}\right)=x\\\\ m\left(\equiv\widehat{CRM}\right)=y\end{array}\right\|$ . Observe that $\boxed{x+y=90^{\circ}+\frac A2}\ (*)$ . Apply the Menelaus' theorem to the transversal $\overline{CRI}$ and $\triangle BMP\ :$

$\frac {PI}{PB}\cdot \frac {MB}{MC}\cdot\frac {RC}{RI}=1\ \stackrel{(*)}{\implies}\ \boxed{IP=IR}$ . Otherwise. Apply the theorem of Sines in the triangles $:\ \ \left\{\begin{array}{cc} \triangle PMB\ : & \frac {BM}{\sin\widehat{BPM}}=\frac {PB}{\sin\widehat{PMB}}\\\\ \triangle RMC\ : & \frac {CM}{\sin\widehat{CRM}}=\frac {RC}{\sin\widehat{RMC}}\end{array}\right\|$ $\implies$ $\sin\widehat{BPM}=\sin\widehat{CRM}\implies$

$\sin x=\sin y\ \stackrel{(*)}{\implies}\ x=y\implies$ $IP=IR$ . Suppose that $PB=RC$ and $b\ne c$ . Thus, $\left\{\begin{array}{ccccc} \widehat{PBA}\equiv\widehat{PBC} & \iff & \frac {PM}{AM}=\frac {BM}{AB+BM} & \iff & \frac {PM}{AM}=\frac a{a+2c}\\\\ \widehat{RCA}\equiv\widehat{RCB} & \iff & \frac {RM}{AM}=\frac {CM}{AC+CM} & \iff & \frac {RM}{AM}=\frac a{a+2b}\end{array}\right\|$ $\implies$

$\boxed{\frac {MP}{MR}=\frac {a+2b}{a+2c}}\ (1)$ . Apply the theorem of Sines in the triangles $:\ \left\{\begin{array}{ccccc} \triangle PMB\ : & \frac {PM}{\sin\frac B2}=\frac {PB}{\sin\widehat{PMB}}\\\\ \triangle RMC\ : & \frac {RM}{\sin\frac C2}=\frac {RC}{\sin\widehat{RMC}}\end{array}\right\|\ \stackrel{PB=RC}{\implies}\ \boxed{\frac {PM}{RM}=\frac {\sin\frac B2}{\sin\frac C2}}\ (2)$ . From the relations $(1)$ and $(2)$

obtain that $\frac {a+2b}{a+2c}=\frac {\sin\frac B2}{\sin\frac C2}$ , i.e. $\frac {a+2b}{a+2c}=\sqrt{\frac {b(s-c)}{c(s-b)}}\implies$ $\frac {c(a+2b)^2}{s-c}=\frac {b(a+2c)^2}{s-b}=$ $\frac {a^2(b+c)+8abc+4bc(b+c)}a=\frac {\left(4bc-a^2\right)(b-c)}{b-c}\ \stackrel{b\ne c}{\implies}$

$a^2(b+c)+8abc+4bc(b+c)=4abc-a^3\implies$ $a^2(a+b+c)+4abc+4bc(b+c)=0$ what isn't truly. In conclusion, $b=c$ , i.e. $AB=AC$ .

PP9 (Cr. Tello). Let an $A$ -right $\triangle ABC$ and $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ so that $\left\{\begin{array}{ccc} BF=CE=s-a\\\\ AF=CD=s-b\\\\ BD=AE=s-c\end{array}\right\|$ , i.e. $\left\{\begin{array}{c} D\in BC\cap w_a\\\\ E\in CA\cap w_b\\\\ F\in AB\cap w_c\end{array}\right\|$ , where $w_a$ is the $A$ -excircle

$\mathbb C\left(I_a,r_a\right)$ a.s.o. Denote $\left|\begin{array}{c} X\in (BC)\ ,\ FX\perp BC\ ;\ FX=x\\\\ Y\in (BC)\ ,\ EX\perp BC\ ;\ EY=y\end{array}\right|\ \wedge\ \left|\begin{array}{ccc} M\in (AB)\ ,\ DM\perp AB\ ;\ DM=m\\\\ N\in (AC)\ ,\ DN\perp AC\ ;\ DN=n\end{array}\right|$ . Prove that $\boxed{x+y+m+n=a}\ (*)$ , where $BC=a$ .

Proof. Denote $\left|\begin{array}{ccc} P\in BC\ ,\ U\in I_cP\ ;\ FU\parallel BC\\\\ R\in BC\ ,\ V\in I_bR\ ;\ EV\parallel BC\end{array}\right|$ . Observe that $\triangle BXF\equiv\triangle EYC\sim$ $\triangle BAC$ $\implies$ $BX=y$ , $CV=x$ and $\frac yc=\frac xb=$ $\frac {s-a}a=\frac {x+y}{b+c}\implies$

$\boxed{x+y=\frac {(s-a)(b+c)}a}\implies$ $\left\{\begin{array}{ccccc} \triangle FUI_c\equiv\triangle CND\implies ND=UI_c=n & \implies & r_c=I_cP=n+x & \implies & n+x=s-b\\\\ \triangle EVI_b\equiv\triangle BMD\implies MD=VI_b=m & \implies & r_b=I_bR=m+y & \implies & m+y=s-c\end{array}\right\|$ $\bigoplus$ $\implies$ the relation $(*)$ .

PP10. Let $\triangle ABC$ and the $B$ -bisector $[AD$ . For $\triangle DBA$ denote its incircle $C\left(I_1,r_1\right)$ and its $B$ -excircle $C\left(J_1,R_1\right)$ .

For $\triangle DBC$ denote its incircle $C\left(I_2,r_2\right)$ and its $B$ -excircle $C\left(J_2,R_2\right)$ . Prove that $\frac {r_1R_1}{r_2R_2}=\frac ca$ (standard notations).

Proof. The incircle and the $B$ -excircle of $\triangle DBA$ touch the sideline $AB$ at $X_1$ , $Y_1$ respectively. The incircle and the $B$ -excircle of $\triangle DBC$ touch the sideline $AB$ at $X_2$ , $Y_2$ respectively.

Let $\left\{\begin{array}{cc} BA=c\ ; & BC=a\\\\ DA=x\ ; & DC=y\end{array}\right\|$ , where $x+y=b\ ,\ \frac xc=\frac ya=\frac b{a+c}$ and $AD=\boxed{l^2=ac-xy}$ . Therefore, $\left\{\begin{array}{ccc} AX_1=c+x-l & ; & AY_1=x+l-c\\\\ CX_2=a+y-l & ; & CY_2=y+l-a\end{array}\right\|$ . Thus,

$\left\{\begin{array}{ccccc} \triangle AX_1I_1\sim J_1Y_1A & \implies & \frac {AX_1}{J_1Y_1}=\frac {X_1I_1}{Y_1A} & \implies & r_1R_1=AX_1\cdot AY_1\\\\ \triangle CX_2I_2\sim J_2Y_2C & \implies & \frac {CX_2}{J_2Y_2}=\frac {X_2I_2}{Y_2C} & \implies & r_2R_2=CX_2\cdot CY_2\end{array}\right\|$ . In conclusion, $\frac {r_1R_1}{r_2R_2}=$ $\frac {AX_1\cdot AY_1}{CX_2\cdot CY_2}=$ $\frac {(c+x-l)(x+l-c)}{(a+y-l)(y+l-a)}=$ $\frac {-l^2+2cl+x^2-c^2}{-l^2+2al+y^2-a^2}=$

$\frac {xy-ac+2cl+x^2-c^2}{xy-ac+2al+y^2-a^2}=$ $\frac {ax-ac+2cl-c^2}{ay-ac+2al-a^2}=$ $\frac {cy-ac+2cl-c^2}{ay-ac+2al-a^2}=$ $\frac {c(y-a+2l-c)}{a(y-c+2l-a)}=\frac ca$ $\implies$ $\frac {r_1R_1}{r_2R_2}=\frac ca$ .

Remark. Prove easily that the area $S$ of $\triangle ABC$ is given by the identity $\boxed {S=rr_a\cot\frac A2}$ . Thus, $\frac {r_1R_1}{r_2R_2}=\frac {r_1R_1\cot\frac B2}{r_2R_2\cot\frac B2}=\frac {[DBA]}{[DBC]}=\frac {DA}{DC}=\frac ca$ .

PP11. Let a square $ABCD$ and $\left\{\begin{array}{cc} N\in (BD)\ ;\ A\in (MC)\\\\ m\left(\widehat{BCN}\right)=m\left(\widehat{ABM}\right)=\phi\in\left(0^{\circ},45^{\circ}\right)\end{array}\right\|$ . Suppose that the circumcircles of $\triangle DNC$ and $\triangle ABM$ are tangent. Find the value of $\phi$ .

Proof 1. Suppose w.l.o.g. that $AB=1$ . The $\triangle CDN$ is acute and let its circumcircle $w_1=C(O_1,R_1)$ . The $\triangle ABM$ is obtuse and let its circumcircle $w_2=C(O_2,R_2)$ , where

$\left\{O_1, O_2\right\}\subset d$ , where $d$ is the bisector of $[AB]$ . Prove easily that $2R_1\sin\left(45^{\circ}+\phi\right)=2R_2\sin \left(45^{\circ}-\phi\right)=1$ and $\left\{\begin{array}{cccc} d_1=\delta_{DC}\left(O_1\right)=R_1\cos\left(45^{\circ}+\phi\right)\\\\ d_2=\delta_{AB}\left(O_2\right)=R_2\cos\left(45^{\circ}-\phi\right)\end{array}\right\|$ . Thus,

$O_1O_2=1+d_2-d_1=$ $1+R_2\cos\left(45^{\circ}-\phi\right)-R_1\cos\left(45^{\circ}+\phi\right)=$ $1+\frac 12\cdot\left[\cot\left(45^{\circ}-\phi\right)-\cot\left(45^{\circ}+\phi\right)\right]=$ $1+\frac 12\cdot \frac {\sin 2\phi}{\sin\left(45^{\circ}-\phi\right)\sin\left(45^{\circ}+\phi\right)}=$ $1+\frac {\sin 2\phi}{\cos2\phi}\implies$

$\boxed{O_1O_2=1+\tan 2\phi}\ (1)$ . Hence $R_1+R_2=\frac 12\cdot\left[\frac 1{\sin \left(45^{\circ}+\phi\right)}+\frac 1{\sin \left(45^{\circ}-\phi\right)}\right]=$ $\frac {\sqrt 2\cos\phi}{\sin \left(45^{\circ}+\phi\right)\sin \left(45^{\circ}-\phi\right)}\implies$ $\boxed{R_1+R_2=\frac {\sqrt 2\cos\phi}{\cos 2\phi}}\ (2)$ . The circles $w_1$ and $w_2$

are tangent $\iff$ $O_1O_2=R_1+R_2\iff$ $1+\tan 2\phi =\frac {\sqrt 2\cos \phi}{\cos 2\phi}\iff$ $\cos 2\phi +\sin 2\phi =\sqrt 2\cos\phi\iff$ $\sin\left(45^{\circ}+2\phi\right)=\cos\phi\iff$ $45^{\circ}-2\phi =\phi\iff$ $\boxed{\phi =\frac {\pi}{12}}$ .

Proof 2 (Sunken Rock). Denote the midpoints $P$ , $Q$ of the sides $[AB]$ , $[CD]$ which share the same perpendicular bisector $PQ$ . So $T\in PQ$ and it belongs to the bisector of $\widehat{BMA}$

with $m\left(\widehat{BMA}\right)=45^{\circ}-\phi\implies$ $m\left(\widehat{PAT}\right)=\frac {45^{\circ}-\phi}{2}$ . Hence $m\left(\widehat{DNC}\right)=45^{\circ}+\phi =$ $m\left(\widehat{DTC}\right)\implies$ $m\left(\widehat{DTQ}\right)=\frac {45^{\circ}+\phi}{2}$ and $m\left(\widehat{TDQ}\right)= 45^{\circ}+\frac {45^{\circ}-\phi}{2}$ .

Suppose $AB=2$ and call $y=\frac {45^{\circ}-\phi}{2}\implies$ $\left\{\begin{array}{ccc} \triangle PAT & \implies & \tan y=PT\\\\ \triangle DTQ & \implies & \tan \left(45^{\circ}+y\right)=TQ\end{array}\right\|$ $\implies$ $\tan y+\tan \left(45^{\circ}+y\right)=PT+TQ=2\stackrel{(\tan y=z)}{\implies}\ z+\frac {1+z}{1-z}=2\implies$

$z^2-4z+1=0\implies$ $z=2-\sqrt 3\implies y=15^{\circ}\implies \phi =15^{\circ}$ .

PP12 (Extension). Let $ABC$ be a triangle and consider the points $D\in BC$ , $E\in CA$ , $F\in AB$ so that the perpendicular

lines from $A$ , $B$ , $C$ on $EF$ , $FD$ , $DE$ are concurrently in the point $P$ (the triangles $ABC$ , $DEF$ are orthologically !).

Define points $\left\|\begin{array}{c} \{A_1,A_2\}\subset BC\ ,\ DA_1 = DA_2 = DP \\ \\ \{B_1,B_2\}\subset CA\ ,\ EA_1 = EA_2 = EP \\ \\ \{C_1,C_2\}\subset AB\ ,\ FA_1 = FA_2 = FP\end{array}\right\|$ . Prove that the points $\{A_1, A_2,B_1,B_2,C_1,C_2\}$ are concyclic.

Remark. See here a particular case - question 1 from day 1 at IMO 2008. Another particular case is

when $P$ is the incenter and $D$ , $E$ , $F$ are the tangent points of incircle with sides of the triangle $ABC$ .

Proof. $AP\perp EF$ $\Longleftrightarrow$ $FA^2 - FP^2 = EA^2 - EP^2$ $\Longleftrightarrow$ the point $A$ has same power w.r.t. the circles $C(F,FP)$ and $C(E,EP)$ $\Longleftrightarrow$

$\overline {AC_1}\cdot\overline{AC_2} = \overline{AB_1}\cdot\overline{AB_2}$ $\Longleftrightarrow$ the quadrilateral $B_1B_2C_1C_2$ is cyclically. Pove analogously the quadrilaterals $C_1C_2A_1A_2$ and $A_1A_2B_1B_2$

are cyclically. Since the radical axis $A_1A_2$ , $B_1B_2$ , $C_1C_2$ aren't concurrently obtain that the points $A_1,A_2,B_1,B_2,C_1,C_2$ are concyclically.

PP13. Let $ABC$ be a triangle and consider the points $D\in BC$ , $E\in CA$ , $F\in AB$ so that the perpendicular lines from $A$ , $B$ , $C$ on $EF$ , $FD$ , $DE$ are concurrently in the point $P$

(the triangles $ABC$ , $DEF$ are orthologically !). Define points $\left\|\begin{array}{c} \{A_1,A_2\}\subset BC\ ,\ DA_1 = DA_2 = DP \\ \\ \{B_1,B_2\}\subset CA\ ,\ EA_1 = EA_2 = EP \\ \\ \{C_1,C_2\}\subset AB\ ,\ FA_1 = FA_2 = FP\end{array}\right\|$ . Prove that the points $\{A_1, A_2,B_1,B_2,C_1,C_2\}$ are concyclic.

Proof. $AP\perp EF$ $\Longleftrightarrow$ $FA^2 - FP^2 = EA^2 - EP^2$ $\Longleftrightarrow$ $A$ has same power w.r.t. $C(F,FP)$ and $C(E,EP)$ $\Longleftrightarrow$ $\overline {AC_1}\cdot\overline{AC_2} = \overline{AB_1}\cdot\overline{AB_2}$ $\Longleftrightarrow$ $B_1B_2C_1C_2$ is cyclic. Prove

analogously that $C_1C_2A_1A_2$ , $A_1A_2B_1B_2$ are cyclic. Since the radical axis $A_1A_2$ , $B_1B_2$ , $C_1C_2$ aren't concurrently obtain that $A_1,A_2,B_1,B_2,C_1,C_2$ are concyclic. Remark. See here

a particular case - question 1 from day 1 at IMO 2008. Another particular case is when $P$ is the incenter and $D$ , $E$ , $F$ are the tangent points of incircle with sides of the triangle $ABC$ .

PP14. Let an $A$ -isosceles $\triangle ABC$ with the diameter $[AD]$ of its circumcircle. For $M$ of $(BC)$ let $MNAP$ be the parallelogram with $N\in (CA)$ and $P\in (AB)$ . Prove that $DM\perp NP$

Proof. Denote $\left\{\begin{array}{c} U\in MN\cap BD\\\\ V\in MP\cap CD\end{array}\right\|$ , Thus, $\left\{\begin{array}{ccc} MN\parallel AB\ ,\ AB\perp BD & \implies & MN\perp BD\\\\ MP\parallel AC\ ,\ AC\perp CD & \implies & MP\perp CD\end{array}\right\|$ . Denote $\left\{\begin{array}{cccc} AP=MN=NC=a\\\\ AN=PM=PB=b\\\\ BD=CD=c\end{array}\right\|$ . Observe that

$\left\{\begin{array}{ccccc} DN^2 & = & CN^2+CD^2 & = & a^2+c^2\\\\ DP^2 & = & BP^2+BD^2 & = & b^2+c^2\end{array}\right\|$ . In conclusion, $DM\perp NP\iff$ $DN^2-DP^2=MN^2-MP^2\iff$ $\left(a^2+c^2\right)-\left(b^2+c^2\right)=a^2-b^2$ what is truly.

PP15. Let $\triangle ABC$ with the incircle $w(I)$ which touches $\triangle ABC$ at $D\in (BC)$ , $E\in (CA)$ and $F\in (AB)$ . Is well-known

that $\Gamma\in AD\cap BE\cap CF$ is the Gergonne's point. Prove that $\Gamma EAF$ is cyclically $\iff \frac 2{s-a}=\frac 1{s-b}+\frac 1{s-c}\iff \frac {\Gamma A}{\Gamma D}=2$ .

Proof 1 (metric). Let $\left\{\begin{array}{ccccc} AE=AF & = & s-a & = & x\\\\ BF=BD & = & s-b & = & y\\\\ CD=CE & = & s-c & = & z\end{array}\right\|$ where $x+y+z=s$ . Apply the Stewart's theorem to the cevian $BE\ :$

$\boxed{(x+z)BE^2+xz(x+z)=z(x+y)^2+x(y+z)^2}\ (1)$ . Apply the Van Aubel's relation $:\ \frac {\Gamma B}{\Gamma E}=\frac {FB}{FA}+\frac {DB}{DC}\iff$ $\frac {\Gamma B}{\Gamma E}=\frac yx+\frac yz\iff$

$\boxed{\frac {\Gamma B}{y(x+z)}=\frac {\Gamma E}{xz}=\frac {BE}{xy+yz+zx}}\ (2)$ . Thus, $\Gamma EAF$ is cyclic $\iff B\Gamma \cdot BE=$ $BF\cdot BA\ \stackrel{(2)}{\iff}\ \frac {y(x+z)\cdot BE^2}{xy+yz+zx}=$ $y(x+y)\ \stackrel{(1)}{\iff}$ $z(x+y)^2+x(y+z)^2=$

$xz(x+z)+(x+y)(xy+yz+zx)\iff$ $x(y+z)^2=xz(x+z)+xy(x+y)\iff$ $(y+z)^2=z(x+z)+y(x+y)\iff$ $2yz=x(y+z)\iff$ $\frac 2x=\frac 1y+\frac 1z$ .

Proof 2 (metric). Find easily that $\left\{\begin{array}{ccc} (x+z)BE^2 & = & z(x+y)^2+x(y+z)^2-xz(x+z)\\\\ (x+y)CF^2 & = & y(x+z)^2+x(y+z)^2-xy(x+y)\end{array}\right\|$ . Observe that $AC\cdot BE\cdot\sin\widehat{AEB}=2S=AB\cdot CF\cdot \sin\widehat {BFC}$

where $S$ is the area of $\triangle ABC$ . Thus, $\Gamma EAF$ is cyclic $\iff \widehat{AEB}\equiv\widehat {BFC}\iff$ $AC\cdot BE=AB\cdot CF\iff$ $(x+z)\cdot (x+z)BE^2=(x+y)\cdot (x+y)CF^2\iff$

$(x+z)\cdot \left[z(x+y)^2+x(y+z)^2-xz(x+z)\right]=(x+y)\cdot\left[y(x+z)^2+x(y+z)^2-xy(x+y)\right]\iff$

$z(x+z)(x+y)^2+x(x+z)(y+z)^2-xz(x+z)^2=y(x+y)(x+z)^2+x(x+y)(y+z)^2-xy(x+y)^2\iff$

$x(x+y)(x+z)(z-y)+x(z-y)(y+z)^2=x(z-y)\left[x^2+2x(y+z)+\left(y^2+yz+z^2\right)\right]\iff$

$(x+y)(x+z)+(y+z)^2=x^2+2x(y+z)+\left(y^2+yz+z^2\right)\iff$ $2yz=x(y+z)\iff$ $\frac 2x=\frac 1y+\frac 1z$ .

Remark. Denote $L\in EF\cap BC$ . Is well-known or can prove easily that $LI\perp AD$ . Denote $K\in LI\cap AD$ . Since

$(B,C,D,L)$ is an harmonical division and $KD\perp KL$ obtain that the ray $[KD$ is the bisector of the angle $\widehat{BKC}\ .$

Lemma. $(*)\ .$ Prove that in $A$ -right $\triangle ABC$ there is the relation $\frac 1{AB^2}+\frac 1{AC^2}=\frac 1{AD^2}\ ,$ where $D\in BC\ ,\ AD\perp BC\ .$ Indeed, $AD\cdot BC=AB\cdot AC\ .$

PP16. A line what pass by the vertex $A$ of the square $ABCD$ cut the lines $BC$ and $CD$ in the points $E$ and $F\ .$ Prove that there is the relation $:\ \boxed{\frac{1}{AE^2}+\frac{1}{AF^2}=\frac{1}{AB^2}}\ .$

Method 1. Let $Y\in CD\ ,\ YA\perp EF\ .$ Hence $\triangle ADY\equiv\triangle ABE\implies AY=AE\ .$ Aplly the lemma $(*)$ to $\triangle AYF\ :\ \frac 1{AF^2}+\frac 1{AY^2}=$ $\frac 1{AD^2}\implies$ $\frac 1{AF^2}+\frac 1{AE^2}=\frac 1{AB^2}\ .$

Method 2. Let $X\in BC\ ,\ XA\perp EF\ .$ Hence $\triangle ABX\equiv\triangle ADF\implies AX=AF\ .$ Apply the lemma $(*)$ to $\triangle AXE\ :\ \frac 1{AE^2}+$ $\frac 1{AX^2}=\frac 1{AB^2}\implies$ $\frac 1{AE^2}+\frac 1{AF^2}=\frac 1{AB^2}\ .$

Method 3. $\triangle ABE\sim\triangle FDA\implies$ $\frac {AB}{AE}=\frac {FD}{FA}\implies$ $\frac {AB^2}{AE^2}=\frac {FD^2}{FA^2}=$ $\frac {FA^2-AD^2}{FA^2}=$ $1-\frac {AD^2}{FA^2}\implies$ $\frac {AB^2}{AE^2}+\frac {AD^2}{FA^2}=$ $1\ \stackrel{(AD=AB)}{\implies}\ \frac 1{AE^2}+$ $\frac 1{AF^2}=\frac 1{AB^2}\ .$

Method 4. $AB\parallel FD\implies$ $m\left(\widehat{EAB}\right)=m\left(\widehat{AFD}\right)=x\implies$ $\left\{\begin{array}{c} \cos x=\cos\widehat{EAB}=\frac {AB}{AE}\\\\\ \sin x=\sin\widehat{AFD}=\frac {AD}{AF}\end{array}\right\|\implies$ $1=\cos^2x+\sin^2x=$ $\frac {AB^2}{AE^2}+\frac {AD^2}{FA^2}\stackrel{(AD=AB)}{\implies}\frac 1{AE^2}+$ $\frac 1{AF^2}=\frac 1{AB^2}\ .$

PP17 (Kadir Altintas, Turkey). Let the semicircle $s=\mathbb S(O,R)$ with the diameter $[AB]$ and the circle

$w=\mathbb C(I,r)$ what is tangent to $AB$ at $C$ and interior tangent to $s$ at $D\ .$ Prove that $\frac 2{CD^2}=\frac 1{AC^2}+\frac 1{BC^2}\ .$

Proof 1. Denote $E\in s$ so that $CE\perp AB\ .$ Observe that $\left\{\begin{array}{ccc} ID=IC=r & ; & OI=R-r\\\\ OC=\sqrt{R^2-2Rr} & ; & CA=R+\sqrt{R^2-2Rr}\\\\ CE^2=CA\cdot CB=2Rr & ; & CB=R-\sqrt{R^2-2Rr}\end{array}\right\|\ .$ Apply the Stewart's relation to $CI$ in $\triangle OCD\ :$

$CD^4\cdot IO+CO^2\cdot ID=CI^2\cdot OD+OD\cdot IO\cdot ID\iff$ $CD^2\cdot (R-r)+r\left(\cancel{R^2}-2Rr\right)=$ $\cancel{r^2R}+Rr(\cancel R-\cancel r)\iff$ $CD=\frac {2Rr^2}{R-r}\iff$ $\boxed{\frac 2{CD^2}=\frac {R-r}{Rr^2}}\ (1)\ .$

On other hand $\frac 1{AC^2}+\frac 1{BC^2}=$ $\left(\frac 1{AC}+\frac 1{BC}\right)^2-\frac 2{AC\cdot BC}=$ $\left(\frac {AC+BC}{AC\cdot BC}\right)^2-\frac 2{AC\cdot BC}=$ $\left(\frac {AB}{CE^2}\right)^2-\frac 2{CE^2}=$ $\left(\frac {\cancel{2R}}{\cancel{2R}r}\right)^2-\frac {\cancel 2}{\cancel{2}Rr}=$ $\frac 1{r^2}-\frac 1{Rr}=\frac {R-r}{Rr^2}\implies$

$\boxed{\frac 1{AC^2}+\frac 1{BC^2}=\frac {R-r}{Rr^2}}\ (2)\ .$ From the relations $(1)$ and $(2)$ obtain the required relation $\frac 2{CD^2}=\frac 1{AC^2}+\frac 1{BC^2}\ .$

Proof 2. From the down-lemma $(\Downarrow )$ (well-known or prove easily) obtain that the ray $[DC$ is the common bisector of $\widehat{ADB}\ ,$ $\widehat{ODP}$ where $P$ is the projection of $D$ on $AB\ .$

Therefore, $\widehat{ADO}\equiv\widehat{ODC}\equiv\widehat{CDP}\equiv\widehat{PDB}\ .$ Hence $IC\parallel PD\implies$ $\left\{\begin{array}{ccccc} \frac {IC}{DP}=\frac {OI}{OD} & \iff & \frac r{DP}=\frac {R-r}R & \iff & \boxed{DP=\frac {Rr}{R-r}}\\\\ \frac {IO}{ID}=\frac {CO}{CP} & \iff & \frac {R-r}{r}=\frac {\sqrt{R^2-2Rr}}{CP} & \iff & \boxed{PC=\frac {r\sqrt{R^2-2Rr}}{R-r}} \end{array}\right\|$ $\implies$

$CD^2=PC^2+PD^2=$ $\frac {Rr^2(R-2r)+R^2r^2}{(R-r)^2}=$ $\frac {Rr^2\left[(R-2r)+R\right]}{(R-r)^2}=$ $\frac {2Rr^2\cancel{(R-r)}}{(R-r)\cancel{^2}}=\frac {2Rr^2}{R-r}\implies$ $CD^2=\frac {2Rr^2}{R-r}\implies$ $\boxed{\frac 2{CD^2}=\frac {R-r}{Rr^2}}\ (1)$ a.s.o.

Lemma $(\Downarrow )$ .

$1\blacktriangleright$ Let the circles $w\ ,$ $\alpha$ where $\alpha$ is interior tangent to $w$ at $A$ and a line $d$ what is tangent to $\alpha$ at $D$ where $d\cap w=\{B,C\}\ .$ Then $[AD$ is the interior $A$ -bisector of $\triangle BAC\ .$
$2\blacktriangleright$ Let the circles $w\ ,$ $\beta$ where $\beta$ is exterior tangent to $w$ at $A$ and a line $d$ what is tangent to $\beta$ at $D$ where $d\cap w=\{B,C\}\ .$ Then $[AD$ is the exterior $A$ -bisector of $\triangle BAC\ .$

PP18 (Ahmed Faiz). Let $:\ \triangle ABC$ with circumcircle $w\ ;$ $P\in (AB),$ $Q\in (AC)$ so that $BP=CQ\ ;$ circumcircle $\alpha$ of $\triangle PAQ\ ;$ $\{A,D\}=w\cap \alpha .$ Prove that $DB=DC.$

Proof. Denote $m\left(\widehat{APD}\right)=m\left(\widehat{AQD}\right)=x\ .$ Observe that $\triangle P\underline{\underline{BD}}\ \stackrel{a.s.a}{\equiv}\ \triangle Q\underline{\underline{CD}}$ because $PB=QC\ ,$ $\widehat{DBP}\equiv \widehat{DCQ}$ and $m\left(\widehat{DPB}\right)=m\left(\widehat{DQC}\right)=180^{\circ}-x\ .$

PP19 (Argentina Cono Sur TST 2014). Let $[AD$ be the $A$ -bisector of $\triangle BAC,$ where $D\in (BC)$ and the circumcircle

$w$ of $\triangle ABD.$ Denote $E\in w$ so that $BE\perp AD.$ Prove that $E\in AO,$ where $O$ is the circumcenter of $\triangle ABC.$

Proof. Let $F\in BE\cap AD$ and $H\in BC$ so that $AH\perp BC.$ Thus, $\left\{\begin{array}{cccc} ABDE & \mathrm{is\ cyclic} & \implies & \widehat{DBE}\equiv\widehat{DAE}\\\\ ABHF & \mathrm{is\ cyclic} & \implies & \widehat{DBE}\equiv\widehat{DAH}\end{array}\right\|\implies$

$\widehat{DAE}\equiv\widehat{DAH}.$ Prove easily or is well known that $\widehat{DAH}\equiv\widehat{DAO}.$ In conclusion, $\widehat{DAE}\equiv\widehat{DAO},$ i.e. $E\in AO.$

PP20. Let $B$ -isosceles $\triangle ABC$ with the circumcenter $O\ ,$ the incenter $I$ and $D\in BC$ so that $OD\perp BI.$ Prove that $ID\parallel AC.$

Denote $E\in DO\cap BI$ and the midpoint $M$ of $[AB].$ Observe that $OEMB$ is cyclic and $\widehat{COD}\equiv$ $\widehat{MOE}\equiv\widehat {MBE}\equiv$ $\widehat{DBE}\implies$ $\widehat{COD}\equiv$ $\widehat{DBE}\implies$ $ODBI$

is cyclic $\implies$ $\widehat{OID}\equiv \widehat{OBD},$ i.e. $m\left(\widehat{OID}\right)=$ $m\left(\widehat{OBD}\right)=$ $B-\left(90^{\circ}-C\right)=$ $B+C-90^{\circ}=$ $90^{\circ}-A=$ $m\left(\widehat{ACM}\right)$ $\implies$ $\widehat{OID}\equiv\widehat{ACM}$ $\implies$ $ID\parallel AC.$

This post has been edited 279 times. Last edited by Virgil Nicula, Sep 8, 2016, 4:57 PM

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Hello teacher, with respect to PP15, I proposed this problem on India Marathon before ^^
See Problem 20 $\Gamma EAF$ is cyclical $\Longleftrightarrow 2D\Gamma =\Gamma A$ , By Van Aubel's Theorem is exactly this.
And this Lemma solve this, beacuse $\Gamma EAF$ is cyclical $\Longleftrightarrow B$ and $C$ are conjugate points WRT $\odot (EAF)$

Kind Regards

by drmzjoseph, May 27, 2015, 12:21 AM

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Thank you. I saw and like very much your proof.

This post has been edited 1 time. Last edited by Virgil Nicula, May 27, 2015, 1:04 AM

by Virgil Nicula, May 27, 2015, 1:04 AM

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dam bro
impressive

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