Let $ABC$ be a triangle such that $\angle BAC = 90^{\circ}$ and $AB = AC$. Let $M$ be the midpoint of $BC$. A point $D \neq A$ is chosen on the semicircle with diameter $BC$ that contains $A$. The circumcircle of triangle $DAM$ cuts lines $DB$ and $DC$ at $E$ and $F$ respectively. Show that $BE = CF$.
Problem
Source: Iberoamerican 2018 Problem 2
Tags: geometry, circumcircle
Tsukuyomi
26.09.2018 19:07
Let $N$ be the second intersection of $\odot{(ADM)}$ and $BC$. Since $\angle{NDA}=\angle{AMN}=\angle{BDC}=90^{\circ},$ we have $$\angle{BDN}=\angle{CDA}=\angle{CBA}=45^{\circ}\implies \angle{BDN}=\angle{NDC}=45^{\circ}.$$So by the angle bisector theorem we have $\dfrac{BD}{CD}=\dfrac{BN}{CN}$. By power of a point we get $$BE=\dfrac{BN\cdot BM}{BD}=\dfrac{CN\cdot CM}{CD}=CF,$$as desired.
GeoMetrix
27.12.2019 08:50
Easy for a #2. The point $M$ was redundant and this infact holds for any point $M$ on $BC$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -15.24, xmax = 15.24, ymin = -9.55, ymax = 10.01; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((-2.16,1.43), 6.020830507496453), linewidth(1.2)); draw((-8.18,1.33)--(3.86,1.53), linewidth(1.2) + dtsfsf); draw((3.86,1.53)--(-2.26,7.45), linewidth(1.2) + dtsfsf); draw((-2.26,7.45)--(-8.18,1.33), linewidth(1.2) + dtsfsf); draw(circle((-1.3918694638253146,4.453590208242104), 3.1196349574880617), linewidth(1.2)); draw((0.8011476943889559,6.672328140436743)--(-8.18,1.33), linewidth(1.2)); draw((0.8011476943889559,6.672328140436743)--(3.86,1.53), linewidth(1.2)); draw((-2.26,7.45)--(-4.38827925558321,3.5854596720674197), linewidth(1.2)); draw((-2.26,7.45)--(1.6045403279325856,5.321720744416781), linewidth(1.2)); /* dots and labels */ dot((-8.18,1.33),dotstyle); label("$B$", (-8.1,1.53), NE * labelscalefactor); dot((3.86,1.53),linewidth(4pt) + dotstyle); label("$C$", (3.94,1.69), NE * labelscalefactor); dot((-2.26,7.45),linewidth(4pt) + dotstyle); label("A", (-2.18,7.61), NE * labelscalefactor); dot((-2.16,1.43),linewidth(4pt) + dotstyle); label("$M$", (-2.08,1.59), NE * labelscalefactor); dot((0.8011476943889559,6.672328140436743),dotstyle); label("D", (0.88,6.87), NE * labelscalefactor); dot((-4.38827925558321,3.5854596720674197),linewidth(4pt) + dotstyle); label("E", (-4.3,3.75), NE * labelscalefactor); dot((1.6045403279325856,5.321720744416781),linewidth(4pt) + dotstyle); label("F", (1.68,5.49), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that there exists a spiral similiarity centred at $A$ such that $BE \mapsto FC$. Note that under this spiral similiarity we also have $EF \mapsto BC \implies AE=AF \implies BE=CF$ by the previous spiral similiarity as desired.$\blacksquare$.
Soundricio
27.12.2019 19:19
Just notice that AB=AC,angleABD=angleACD and angleAFD=angleAED so angle CAF=angle BAF. Hence the triangles BAF and AFC are congruent and the result follows.
Qzproover
28.12.2019 09:49
Easy angle chasing shows that triangles BEM and MFC are equal ,so BE=FC
trying_to_solve_br
27.07.2021 21:15
mehhhhhhhhhhhhhhhhhhhhhhhhhhhhhh My solution was pretty big but it was nice though Here, let $X=(DAM)$. Let $AB$ touch $X$ in $G$, and $AC$ touch $X$ in $H$. First, notice that: $BG/BD=BE/BA=BE/CA$ and we want $BE/AC=CF/CA=CH/CD$, hence we need $CH/CD=BG/BD$. So that means we also need to prove there's a spiral similarity $\phi: BG \to CH$ $(i)$, and call $\angle MBD=a$. So, as the triangle is isosceles and because of cyclic quads we have $\angle FCA=45+a$, $\angle AMD=90+2a=180-\angle DHA=180-\angle DHC=90-2a$, proving $DH=CH$! So to prove $(i)$, we just need $GD=GB$. In other words, we just need $\angle ADG=a$, so the quadrilateral $ADHG$ must be an isosceles trapezoid. Now we focus on proving $\angle GDF=45-a$ $(ii)$. Thus, one may notice $MH$ is the perpendicular bisector of $CD$ (it is easy to see because $MD=MC$ and $HD=HC$). Then $\angle AFC=90-2a=2(\angle MHA)$. So $FA=FC$, and then $\angle HAG=90=\angle FAH+\angle FAG=45+a + \angle FAG$. Finally, $\angle GAF=45-a$, proving $(ii)$.
Math_Is_Fun_101
02.10.2021 21:18
We use directed angles$\mod 180^\circ$. Then, the cyclic quads give \begin{align*} \measuredangle ABE = \measuredangle ABD = \measuredangle ACD = \measuredangle ACF \\ \measuredangle BEA = \measuredangle DEA = \measuredangle DFA = \measuredangle CFA, \end{align*}implying $\triangle EAB\stackrel+\sim\triangle FAC$. Since $AB=AC$, the triangles are in fact congruent, so $BE=CF$. $\blacksquare$ Remark: A lot of things about this problem are unnecessary. Indeed, $\triangle ABC$ need not be right and $M$ need not be the midpoint of $\overline{BC}$. The following generalization holds and is proven in the exact same way as above: Let $ABC$ be a triangle with $AB=AC$ and circumcircle $\Gamma$. Let $D$ be a point on arc $\widehat{BC}$ of $\Gamma$ containing $A$ and $\omega$ be any circle passing through $A$ and $D$. If lines $\overline{DB}$ and $\overline{DC}$ meet $\omega$ again at $E$ and $F$ respectively, prove that $BE=CF$.
ilikemath40
03.10.2021 18:44
[asy][asy]
size(10cm);
import olympiad;
pair A, B, C, D, E, F, G, M;
B = (0, 0);
C = (8, 0);
A = (4, 4);
draw(A--B);
draw(B--C);
draw(A--C);
M = (4, 0);
draw(circle((4, 0), sqrt(16)));
D = (5.96, 3.48);
draw(circumcircle(A, D, M));
draw(D--B);
draw(D--C);
draw(A--D);
E = (2.52, 1.48);
F = (6.52, 2.52);
draw(E--F);
draw(A--E);
draw(A--F);
draw(A--B);
G = (2, 2);
draw(G--E);
draw(A--M);
draw(M--D);
dot(A^^B^^C^^D^^E^^F^^G^^M);
label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, dir(20));
label("$D$", D, NE);
label("$E$", E, S);
label("$F$", F, NE);
label("$G$", G, N);
label("$M$", M, S);
[/asy][/asy]
For the sake of my sanity, call $\angle{EBG}=\alpha$. Also let $G$ be the midpoint of $AB$. Then we do a bit of angle chasing.
Because $\angle{DCA}=\alpha$, we have $\angle{DMA}=\angle{DEA}=2\alpha$. Also notice that since $\angle{ACB}=45^\circ$, $\angle{ADB}=\angle{ADE}=45^\circ$ as well. Notice $\angle{CBD}=45-\alpha$ and $\angle{DCB}=45+\alpha$.
Claim 1. $EF$ is diameter of $(ADM)$
Proof. Because $\angle{BDC} = 90^\circ$, $\angle{EDF}=90^\circ$. $\blacksquare$
Claim 2. $\angle{FED}=45-2\alpha$, $\angle{DFE}=45+2\alpha$
Proof. Because Claim 1, we have $\angle{EDF}=90^\circ$ we have \[ \angle{ADF}=\angle{ADE}+\angle{EDF}=135^\circ. \]Now considering the cyclic quadrilateral $ADFE$ we know that \[ \angle{FEA}=\angle{DEA}+\angle{FED}=2\alpha+\angle{FED}=45^\circ \]where the last equality is from $\angle{ADF}+\angle{FEA}=180^\circ$. So we know that $\angle{FED}=45-2\alpha$. This also shows that $\angle{DFE}=45+2\alpha$ because of Claim 1. $\blacksquare$
Claim 3. $\frac{BC}{EF}=2\cos(\alpha)$
Proof. Consider $\triangle{AEF}$. It is a $45-45-90$ triangle and is similar to $\triangle{ABC}$. Furthermore, we know that \[ \angle{AEB}=180-\angle{DEA}=180-2\alpha. \]This means that $\angle{BAE}=\alpha$ implying $\triangle{AEB}$ is isosceles. Since $G$ is the midpoint of $AB$ we know that $\angle{BGE}=90^\circ$. Then $\frac{BG}{EB}=\cos(\alpha)$ and since \[ \frac{BG}{EB}=\frac{AB}{2AE}=\cos(\alpha) \implies \frac{AB}{AE}=2\cos(\alpha). \]Since $\triangle{AEF}\sim \triangle{ABC}$ we also know that $\frac{BC}{EF}=2\cos(\alpha).$ $\blacksquare$
Claim 4. $BE=CF$
Proof. \begin{align*}
BD&=BC\sin(45+\alpha) \\
ED&=EF\sin(45+2\alpha) \\
CD&=BC\sin(45-\alpha) \\
FD&=EF\sin(45-2\alpha)
\end{align*}Now
\begin{align*}
BE&=BD-ED \\
&=BC\sin(45+\alpha)-EF\sin(45+2\alpha) \\
&=\frac{\sqrt{2}}{2}BC(\sin(\alpha)+\cos(\alpha))-\frac{\sqrt{2}}{2}EF(\sin(2\alpha)+\cos(2\alpha))
\end{align*}And
\begin{align*}
CF&=CD-FD \\
&=BC\sin(45-\alpha)-EF\sin(45-2\alpha) \\
&= \frac{\sqrt{2}}{2}BC(\cos(\alpha)-\sin(\alpha))-\frac{\sqrt{2}}{2}EF(\cos(2\alpha)-\sin(2\alpha))
\end{align*}
If we set these two to be equal to each other we get
\begin{align*}
BC(\sin(\alpha)+\cos(\alpha))-EF(\sin(2\alpha)+\cos(2\alpha))&=BC(\cos(\alpha)-\sin(\alpha))-EF(\cos(2\alpha)-\sin(2\alpha)) \\
BC(-2\sin(\alpha))&=EF(-2\sin(2\alpha)) \\
\frac{BC}{EF}&=2\cos(\alpha)
\end{align*}which we have already proved in Claim 3 and thus completing our proof. $\blacksquare$
[asy][asy]
size(10cm);
import olympiad;
pair A, B, C, D, E, F, G, M;
B = (0, 0);
C = (8, 0);
A = (4, 4);
draw(A--B);
draw(B--C);
draw(A--C);
M = (4, 0);
draw(circle((4, 0), sqrt(16)));
D = (5.96, 3.48);
draw(circumcircle(A, D, M));
draw(D--B);
draw(D--C);
draw(A--D);
E = (2.52, 1.48);
F = (6.52, 2.52);
draw(E--F);
draw(A--E);
draw(A--F);
draw(A--B);
G = (2, 2);
draw(G--E);
draw(A--M);
draw(M--D);
dot(A^^B^^C^^D^^E^^F^^G^^M);
label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, dir(20));
label("$D$", D, NE);
label("$E$", E, S);
label("$F$", F, NE);
label("$G$", G, N);
label("$M$", M, S);
[/asy][/asy]
For the sake of my sanity, call $\measuredangle{EBG}=\alpha$. Also let $G$ be the midpoint of $AB$. Then we do a bit of angle chasing.
Because $\measuredangle{DCA}=\alpha$, we have $\measuredangle{DMA}=\measuredangle{DEA}=2\alpha$. Also notice that since $\measuredangle{ACB}=45^\circ$, $\measuredangle{ADB}=\measuredangle{ADE}=45^\circ$ as well. Notice $\measuredangle{CBD}=45-\alpha$ and $\measuredangle{DCB}=45+\alpha$.
Claim 1. $EF$ is diameter of $(ADM)$
Proof.Because $\measuredangle{BDC} = 90^\circ$, $\measuredangle{EDF}=90^\circ$. $\blacksquare$
Claim 2. $\measuredangle{FED}=45-2\alpha$, $\measuredangle{DFE}=45+2\alpha$
Proof. Because Claim 1, we have $\measuredangle{EDF}=90^\circ$ we have \[ \measuredangle{ADF}=\measuredangle{ADE}+\measuredangle{EDF}=135^\circ. \]Now considering the cyclic quadrilateral $ADFE$ we know that \[ \measuredangle{FEA}=\measuredangle{DEA}+\measuredangle{FED}=2\alpha+\measuredangle{FED}=45^\circ \]where the last equality is from $\measuredangle{ADF}+\measuredangle{FEA}=180^\circ$. So we know that $\measuredangle{FED}=45-2\alpha$. This also shows that $\measuredangle{DFE}=45+2\alpha$ because of Claim 1. $\blacksquare$
Claim 3. $\frac{BC}{EF}=2\cos(\alpha)$
Proof. Consider $\triangle{AEF}$. It is a $45-45-90$ triangle and is similar to $\triangle{ABC}$. Furthermore, we know that \[ \measuredangle{AEB}=180-\measuredangle{DEA}=180-2\alpha. \]This means that $\measuredangle{BAE}=\alpha$ implying $\triangle{AEB}$ is isosceles. Since $G$ is the midpoint of $AB$ we know that $\measuredangle{BGE}=90^\circ$. Then $\frac{BG}{EB}=\cos(\alpha)$ and since \[ \frac{BG}{EB}=\frac{AB}{2AE}=\cos(\alpha) \implies \frac{AB}{AE}=2\cos(\alpha). \]Since $\triangle{AEF}\sim \triangle{ABC}$ we also know that $\frac{BC}{EF}=2\cos(\alpha).$ $\blacksquare$
Claim 4. $BE=CF$
Proof. \begin{align*}
BD&=BC\sin(45+\alpha) \\
ED&=EF\sin(45+2\alpha) \\
CD&=BC\sin(45-\alpha) \\
FD&=EF\sin(45-2\alpha)
\end{align*}Now
\begin{align*}
BE&=BD-ED \\
&=BC\sin(45+\alpha)-EF\sin(45+2\alpha) \\
&=\frac{\sqrt{2}}{2}BC(\sin(\alpha)+\cos(\alpha))-\frac{\sqrt{2}}{2}EF(\sin(2\alpha)+\cos(2\alpha))
\end{align*}And
\begin{align*}
CF&=CD-FD \\
&=BC\sin(45-\alpha)-EF\sin(45-2\alpha) \\
&= \frac{\sqrt{2}}{2}BC(\cos(\alpha)-\sin(\alpha))-\frac{\sqrt{2}}{2}EF(\cos(2\alpha)-\sin(2\alpha))
\end{align*}
If we set these two to be equal to each other we get
\begin{align*}
BC(\sin(\alpha)+\cos(\alpha))-EF(\sin(2\alpha)+\cos(2\alpha))&=BC(\cos(\alpha)-\sin(\alpha))-EF(\cos(2\alpha)-\sin(2\alpha)) \\
BC(-2\sin(\alpha))&=EF(-2\sin(2\alpha)) \\
\frac{BC}{EF}&=2\cos(\alpha)
\end{align*}which we have already proved in Claim 3 and thus completing our proof. $\blacksquare$
ilikemath40
03.10.2021 18:46
trying_to_solve_br wrote: mehhhhhhhhhhhhhhhhhhhhhhhhhhhhhh My solution was pretty big ... Are you sure about that?