Right angle at the midpoint of EF

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Right angle at the midpoint of EF

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Source: Baltic Way 2022/13

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#1 h Nov 12, 2022, 8:15 PM

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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 $\angle BMD = 90^o$ .

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#2 h Nov 12, 2022, 8:25 PM • 3 Y

Y by i_am_from_europe, mkomisarova, Infinityfun

This problem was proposed by me.

Solution

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Reason: typo

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#3 h Nov 12, 2022, 8:33 PM • 1 Y

Y by Infinityfun

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.

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Hermione.Potter

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Hermione.Potter

#4 h Nov 13, 2022, 2:55 PM

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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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#5 h Nov 13, 2022, 4:19 PM

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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.

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#6 h Nov 17, 2022, 3:52 PM

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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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#7 h May 28, 2023, 12:57 PM

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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.

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#8 h Aug 25, 2023, 12:28 PM

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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.

This post has been edited 2 times. Last edited by SerdarBozdag, Aug 25, 2023, 12:30 PM

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