Let $ABCD$ be a cyclic quadrilateral with $AB < BC$ and $AD < DC$. Let $E$ and $F$ be points on the sides $BC$ and $CD$, respectively, such that $AB = BE$ and $AD = DF$. Let further M denote the midpoint of the segment $EF$. Prove that $\angle BMD = 90^o$.
Problem
Source: Baltic Way 2022/13
Tags: geometry
Kimchiks926
12.11.2022 23:25
This problem was proposed by me.
Let $T$ be the reflection of $E$ over $BM$. Note that $BA=BT=BE$ and $ME=MT=MF$, therefore, $B$ is the circumcentre of $\triangle ATE$ and $M$ is the circumcentre of $\triangle ETF \implies \angle ETF=90^{\circ}$.
Since $B$ is the circumcentre of $\triangle ATE$, we have that: $$ \angle ATE =180^{\circ}-0.5 \angle ABE=180^{\circ}-0.5 \angle ABC$$Since $\angle ETF=90^{\circ}$, we can calculate $\angle ATF$. Indeed:
$$ \angle ATF =360^{\circ} -\angle ATE -\angle ETF = 90^{\circ}+0.5 \angle ABC $$On the other hand:
$$\angle ABC +\angle CDA=180^{\circ} \implies 0.5 \angle ABC=90^{\circ}-0.5 \angle CDA $$Therefore:
$$ \angle ATF =90^{\circ}+0.5 \angle ABC =180^{\circ}-0.5 \angle CDA $$By converse properties of the the circumcentre, we can conclude that $D$ is the circumcentre of $\triangle ATF$ (since $AD=DF$ and the established angle condition). Therefore, $DT=DF$, which combined together with $MT=MF$ gives us that $MD \perp TF$. Also note that $BM \perp TE$.
Consider the quadrilateral formed by the lines $BM,MD,ET,TF$. We have established that three of its angles are $90^{\circ}$, therefore, the fourth is $90^{\circ}$ as well, which gives us $\angle BMD=90^{\circ}$, as desired.
VicKmath7
12.11.2022 23:33
Here is a different solution. Let $X$ and $Y$ be the midpoints of $AE$ and $AF$, and let $DM \cap AE = R, BM \cap AF=S$. It is easy to see that triangles $BXM$ and $MYD$ are similar since triangles $ABX$ and $DAY$ are similar due to angle chasing. So, we have that $\angle XBM =\angle YMD= \angle ARD$ (since $YM \parallel AR$ as a midline in $\triangle AEF$), so $BXMR$ is cyclic and $\angle BMR= \angle BXR =90^o$, done.
Hermione.Potter
13.11.2022 17:55
Draw the circle centred at $B$ with radius $AB=BE$ and the circle centred at $D$ with radius $AD=DF$. Denote by $H$ the intersection of the two circles. By angle chase, we have $$\angle FHE = 360^o - \angle AHE - \angle AHF = 360^o - (180^o - \frac{1}{2}\angle ABE) - (180^o - \frac{1}{2}\angle ADF)=\frac{1}{2}(\angle ABE+\angle ADF) = 90^o$$This yields $ME=MF=MH$. Now, quadrilaterals $DFMH$ and $BEMH$ are kites, thus $HF \perp DM$ and $HE \perp BM$ which suffices to show $\angle BMD = 90^o$ since $\angle FHE = 90^o$ as well.
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alchemyst_
13.11.2022 19:19
Let $E'$ be the reflection of $E$ over $B$ and $F'$ be the reflection of $F$ over $D$. Note that $90 = \angle{F'AF} = \angle{EAE'}$ and $\angle{AE'E} = \frac{1}{2}\angle{ABE} = \frac{1}{2}(180 - \angle{ADC}) = \angle{AFF'}$ and so $\bigtriangleup F'AF$ and $\bigtriangleup EAE'$ are similar. By spiral similarity, $\bigtriangleup EAF'$ and $\bigtriangleup E'AF$ are also similar with a rotation of $90$ degrees. Thus, $E'F \perp EF'$. Since $E'F || BM$ and $F'E || DM$ we have $\angle{BMD} = 90 $ as required.
Quidditch
17.11.2022 18:52
Let $X,Y$ be the projection from $D,B$ to $AF,AE$ respectively. Note that $XYM$ is the medial triangle of $AEF$. Claim: $\triangle XMD\sim\triangle YBM$ Proof. First, we have $$\angle DXM=90^{\circ}+\angle FXM=90^{\circ}+\angle FAE=90^{\circ}+\angle MYE=\angle MYB.$$It's suffice to prove that \begin{align*} \frac{XM}{XD}=\frac{YB}{YM} &\iff \frac{YA}{XD}=\frac{YB}{XA} \\ &\iff \frac{YA}{YB}=\frac{XA}{XD} \\ &\iff \tan \angle ABY=\tan\angle DAX \\ &\iff \angle ABY=\angle DAX \\ &\iff \frac{1}{2}\angle ABE=90^{\circ}-\frac{1}{2}\angle ADF \\ &\iff \angle ADF+\angle ABE=180^{\circ} \end{align*}which is true. Now, we have \begin{align*} \angle BMD &=\angle BMY+\angle XMY+\angle DMX \\ &=\angle XMY+(\angle BMY+\angle YBM) \\ &=\angle XMY+(180^{\circ}-\angle BYM) \\ &=\angle MYE+(180^{\circ}-(90^{\circ}+\angle MYE)) \\ &= 90^{\circ}. \end{align*}
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Assassino9931
28.05.2023 15:57
A mixture of the above solutions - minimal number of additional points and no trig. Let $X$ and $Y$ be the midpoints of $AE$ and $AF$, respectively. Then $\angle AYD = \angle AXB = 90^{\circ}$ from $AB = BE$, $AD = DF$ and since $ABCD$ is cyclic, we obtain $\angle ADY = \frac{\angle ADC}{2} = 90^{\circ} - \frac{\angle ABC}{2} = 90^{\circ} - \angle ABX = \angle BAX$. Hence $\triangle ADY \sim \triangle BAX$ and $\frac{DY}{AY} = \frac{AX}{XB}$. However, $MX$ and $MY$ are midsegments in $AEF$, thus $MY = AX$, $MX = AY$ and so $\frac{DY}{MX} = \frac{MY}{XB}$, i.e. $\frac{BX}{MX} = \frac{MY}{DY}$. Together with $\angle FYM = \angle EXM = \angle EAF$ and $\angle BXE = \angle FYD$, we deduce $\triangle BXM \sim \triangle \triangle MYD$. Therefore $\angle BMD = \angle BMX + \angle XMY + \angle YMD = \angle MDY + \angle MYF + \angle YMD = 180^{\circ} - \angle DYF = 90^{\circ}$, as desired.
SerdarBozdag
25.08.2023 15:28
Gorgeous solution to gorgeous problem. Reflect $D$ across the $M$ and name it $D’$. Observe that $D’EB$ and $DAB$ are congruent. Thus, $BD’=BD$ with $MD=MD’$ finishes the proof.