Let $ABCD$ be a cyclic quadrilateral such that $AB=AD$. Let $E$ and $F$ be points on the sides $BC$ and $CD$, respectively, such that $BE+DF=EF$. Prove that $\angle BAD = 2 \angle EAF$.
Problem
Source: 2nd Memorial Mathematical Competition "Aleksandar Blazhevski - Cane" - Problem 1
Tags: geometry, cyclic quadrilateral
djmathman
13.01.2021 01:41
This one's cute too
The crucial claim is that $A$ is the $C$-excenter of $\triangle CEF$. The result follows by a simple angle chase.
Without loss of generality assume $BC < CD$. Let $P$ and $Q$ be the projections of $A$ onto $BC$ and $CD$, respectively. As $AB = AD$ and $\measuredangle ABC = \measuredangle ADC$, right triangles $APB$ and $AQD$ are congruent, whence $AP = AQ$ and $PB = DQ$. The first condition implies there exists a circle $\omega$ centered at $A$ tangent to $AB$ and $AD$ at $P$ and $Q$, respectively. The second condition tells us that
\[
EF = BE + DF = (BE + BP) + (DF - DQ) = EP + FQ,
\]so there exists a point $T$ on $\overline{EF}$ such that $EP = ET$ and $FT = FQ$. But now $\odot(PTQ)$ is the $C$-excircle of $\triangle CEF$, which is tangent to $CE$ and $CF$ at $P$ and $Q$, respectively. So $\omega \equiv \odot(PTQ)$ and hence $A$ is the $C$-excenter of $\triangle CEF$.
Euler365
14.04.2021 05:06
Here is a short proof.
Let $P$ be a be a point on $CB$ s.t. $BP=DF$ and $E-B-P$.
$DF=BP$ and $A$ is midpoint of arc $BD$ not containing $C$. Then it is well known that $AF=AP$ (this follows by showing $\triangle ADF\cong\triangle ABP$). $EP=EB+BP=EB+FD=EF$ and $AF=AP$. Hence $AFEP$ is a kite. So $\angle EAF=\frac{\angle PAF}{2}=\frac{\angle BAD}{2}$.
Q.E.D.
angrybird_r
14.04.2021 07:15
djmathman wrote: This one's cute too
The crucial claim is that $A$ is the $C$-excenter of $\triangle CEF$. The result follows by a simple angle chase.
Without loss of generality assume $BC < CD$. Let $P$ and $Q$ be the projections of $A$ onto $BC$ and $CD$, respectively. As $AB = AD$ and $\measuredangle ABC = \measuredangle ADC$, right triangles $APB$ and $AQD$ are congruent, whence $AP = AQ$ and $PB = DQ$. The first condition implies there exists a circle $\omega$ centered at $A$ tangent to $AB$ and $AD$ at $P$ and $Q$, respectively. The second condition tells us that
\[
EF = BE + DF = (BE + BP) + (DF - DQ) = EP + FQ,
\]so there exists a point $T$ on $\overline{EF}$ such that $EP = ET$ and $FT = FQ$. But now $\odot(PTQ)$ is the $C$-excircle of $\triangle CEF$, which is tangent to $CE$ and $CF$ at $P$ and $Q$, respectively. So $\omega \equiv \odot(PTQ)$ and hence $A$ is the $C$-excenter of $\triangle CEF$.
like u
Mathematicsislovely
25.04.2021 16:44
Let $G$ be the point on $EF$ such that $Df=FG$. For obvious reason,$EG=EB$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.cm); 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 = 2., xmax = 10., ymin = -2., ymax = 6.; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); pen ffttww = rgb(1.,0.2,0.4); pen ffttcc = rgb(1.,0.2,0.8); pen zzqqtt = rgb(0.6,0.,0.2); /* draw figures */ draw(circle((6.0710626698987635,1.8728806204548638), 2.837261929360878), linewidth(0.4) + red); draw(circle((4.2,-0.26), 3.8374221313196606), linewidth(0.4) + dotted + ffqqff); draw((4.2,-0.26)--(8.036445189976266,-0.17341525394219356), linewidth(0.4) + red); draw((8.036445189976266,-0.17341525394219356)--(8.7,2.94), linewidth(0.4) + red); draw((8.7,2.94)--(3.7862531525930767,3.5550520520961135), linewidth(0.4) + ffttww); draw((3.7862531525930767,3.5550520520961135)--(4.2,-0.26), linewidth(0.4) + red); draw((4.2,-0.26)--(8.424729695598426,1.648425058887727), linewidth(0.4) + ffttcc); draw((5.391188734424712,3.3541628012368547)--(8.424729695598426,1.648425058887727), linewidth(0.4) + blue); draw((5.391188734424712,3.3541628012368547)--(4.2,-0.26), linewidth(0.4) + ffqqff); draw((3.7862531525930767,3.5550520520961135)--(6.801051632342072,2.561407276490327), linewidth(0.4) + ffqqff); draw((6.801051632342072,2.561407276490327)--(8.036445189976266,-0.17341525394219356), linewidth(0.4) + ffqqff); draw((6.801051632342072,2.561407276490327)--(4.2,-0.26), linewidth(0.4) + dotted + zzqqtt); /* dots and labels */ dot((4.2,-0.26),linewidth(1.pt) + dotstyle); label("$A$", (3.841233596942836,-0.7112628940746841), NE * labelscalefactor); dot((8.036445189976266,-0.17341525394219356),linewidth(1.pt) + dotstyle); label("$B$", (8.230070340423964,-0.5176377436269876), NE * labelscalefactor); dot((8.7,2.94),linewidth(1.pt) + dotstyle); label("$C$", (8.789431886161754,2.989128870036851), NE * labelscalefactor); dot((3.7862531525930767,3.5550520520961135),linewidth(1.pt) + dotstyle); label("$D$", (3.712150163311038,3.61303213259054), NE * labelscalefactor); dot((8.424729695598426,1.648425058887727),linewidth(1.pt) + dotstyle); label("$E$", (8.595806735714058,1.3540720440340797), NE * labelscalefactor); dot((6.801051632342072,2.561407276490327),linewidth(1.pt) + dotstyle); label("$G$", (6.896208192895386,2.6018785691414577), NE * labelscalefactor); dot((5.391188734424712,3.3541628012368547),linewidth(1.pt) + dotstyle); label("$F$", (5.47629042294561,3.3978930765375437), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Observe that, \begin{align*} \angle DGB&=180^{\circ}-\angle FGD-\angle EGB\\ &=180^{\circ}-\angle FDG-\angle EBG\\ &=\angle ADF+\angle ABE-\angle FDG-\angle EBG\\ &=\angle ADG+\angle ABG \end{align*}Therefore,from quadrilateral $ADGB$ we get,$$\angle DAB=360^{\circ}-\angle DGB-\angle ADG-\angle ABG=-2\angle DGB$$As $AD=AB$ ,$A$ is the center of $\odot(DGB)$. Therefore,$AB=AG=AB$ and so $\angle DAF=\angle FAG$ ,$\angle GAE=\angle EAB$. We are done.
squarc_rs3v2m
11.12.2022 12:11
Let $A'$ be the $C$-excenter of $\triangle CFE$, and let $(F), (E)$ be tangent at $T \in EF$. Then obviously $$\triangle DFA' \cong \triangle TFA', \triangle BEA' \cong \triangle TEA' \implies \measuredangle A'DC = \measuredangle FTA' = \measuredangle A'BE \implies A' = A.$$Then the congruent triangles give that $B, D$ are the reflections of $T$ in $AE, AF$ which finishes.