A convex quadrilateral $ABCD$ has right angles at $A$ and $C$. A point $E$ lies on the extension of the side $AD$ beyond $D$ so that$\angle ABE =\angle ADC$. The point $K$ is symmetric to the point $C$ with respect to point $A$. Prove that$\angle ADB =\angle AKE$ .
Problem
Source: International Olympiad of metropolises 2016
Tags: IOM, geometry
Guendabiaani
08.09.2016 01:51
Let $B'$ be the point symmetric to $B$ with respect to $A$ and let $KB' \cap AD = E'$ then since $E'A \perp BB'$ and $A$ is the midpoint of $BB'$ we have $\angle ABE' = \angle AB'E'$ but since $K,B'$ are symmetric to $C,B$ with respect to $A$ we have $BC \parallel KB'$ so $\angle AB'E' = 180^{\circ} - \angle ABC = \angle ADC = \angle ABE$ so $E'=E$ and since $KE \parallel BC$ we have $\angle ADB = \angle ACB = \angle CKE = \angle AKE$.
Math_CYCR
08.09.2016 02:44
Let $D'$ be the reflection of $D$ through $A$ and let $T =AK \cap BP$. $\angle ADC = \angle ABE= 90- \angle AEB$. Furthermore $\angle BDC= 90- 2 \angle AEB - \angle DBE$. Thus $\angle CAD= \angle CBD= 2 \angle AEB+ \angle DBE$. Easily $BD= BD'$ which implies $\angle BD'D= \angle BDD'= \angle AEB+ \angle DBE$. Thus $\angle D'TA= \angle AEB$. $\Longrightarrow BAET$ is cyclic. And $CDKD'$ is a parallelogram. Thus $\angle KD'A= \angle ADC= \angle ABE= \angle ATE$. Thus $D'TEK$ is cyclic. $\Longrightarrow \angle AKE= \angle AD'B= \angle ADB$
Math_CYCR
08.09.2016 03:04
$\angle ADC = \angle ABE= 90- \angle AEB$. Furthermore $\angle BDC= 90- 2 \angle AEB - \angle DBE$. Thus $\angle CAD= \angle CBD= 2 \angle AEB+ \angle DBE$
By law of sines in $\triangle AKE$ and $\triangle BDE$ we get:
$\frac{BE}{BD} = \frac{ \sin \angle BDA}{ \sin \angle BED}$. and $\frac{AE}{AK} = \frac{ \sin \angle AKE}{ \sin \angle AEK}$.
Thus if we prove $\frac{BE}{BD} = \frac{AE}{AK} = \frac{AE}{AC}$ we are done.
$\Longrightarrow \frac{BE}{AE} = \frac{BD}{AC}$
Law of sines in $\triangle ABE$ gives:
$\frac{BE}{AE} = \frac{1}{ \sin \angle ABE}$
Law of sines in $\triangle ABC$ and $\triangle BDC$ gives:
$\frac{BD}{AC} = \frac{1}{ \sin \angle ABC} = \frac{1}{ \sin \angle ADC}$
Thus:
$\frac{BD}{AC} = \frac{1}{ \sin \angle ADC} = \frac{1}{ \sin \angle ABE} = \frac{BE}{AE}$
Done!
FabrizioFelen
08.09.2016 07:50
Let $X=AD\cap BC$, so from $\measuredangle ABE =\measuredangle ADC$ we get $A$ is midpoint of $XE$, futhermore $A$ is midpoint of $CK$ $\Longrightarrow$ $CX\parallel KE$ hence $\measuredangle AKE=\measuredangle ACB$, so from $ABCD$ is cyclic $\measuredangle ADB=\measuredangle ACB$ $\Longrightarrow$ $\measuredangle ADB=\measuredangle AKE$ .
anantmudgal09
13.09.2016 22:30
Nice! (And quite simple compared to $2,3,5,6$ ) Let $AD$ and $BC$ meet at point $P$. Let $Q$ be the point symmetric to $P$ in $A$. We have $\angle ADC=\angle ABP=\angle ABQ$ and so $Q=E$. Now, since $CPKQ$ is a parallelogram, we see that $\angle ADB=\angle ACB=\angle AKE$ and the result follows.
IndoMathXdZ
10.07.2019 10:33
I got the idea when I drew the diagram the wrong way Let the reflection of $E$ with respect to $A$ be $X$, notice that $\measuredangle ABC = \measuredangle ADC = \measuredangle EBA = \measuredangle ABX$. Therefore, we have $X,B,C$ being collinear. Furthermore, $EA = AX$. Now, notice that $CEKX$ is a parallelogram as $EA = AX$ and $CA = AK$. Therefore, we must have \[ \measuredangle AKE = \measuredangle CKE = \measuredangle KCX \equiv \measuredangle ACB = \measuredangle ADB \]
Pluto1708
15.08.2019 20:53
Wolowizard wrote: A convex quadrilateral $ABCD$ has right angles at $A$ and $C$. A point $E$ lies on the extension of the side $AD$ beyond $D$ so that$\angle ABE =\angle ADC$. The point $K$ is symmetric to the point $C$ with respect to point $A$. Prove that$\angle ADB =\angle AKE$ . Quite a simple problem
Let $F=AD\cap BC$.Let $E'$ be reflection of $F$ over $A$.Note that $\angle ADC=\angle ABF=\angle ABE'\angle ABE \Rightarrow E\equiv E'$.Hence $FKEC$ is a parallelogram so $\angle AKE=\angle ACF=\angle ADB$ as desired
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Jalil_Huseynov
20.09.2021 08:56
Let $F=AD\cap BC$ $\implies$ $AF=AE$ and since $AK=AC$ we get $BC||KE$ $\implies$ $\angle AKE=\angle ACB=\angle BDA$. So done!