Given quadrilateral $ABCD$ inscribed into a circle with diagonal $AC$ as diameter. Let $E$ be a point on segment $BC$ s.t. $\sphericalangle DAC=\sphericalangle EAB$. Point $M$ is midpoint of $CE$. Prove that $BM=DM$.
Problem
Source: Polish MO 2022 P4
Tags: geometry
badumtsss
12.02.2022 17:30
Let $AE\cap (ABCD)=D'$. Then, because $M$ is the midpoint of $CE$, $\angle MD'C = \angle DAC$, so $D,M,D'$ collinear. $\triangle DCM \cong BD'M$ which finishes the proof.
songhongyi
12.02.2022 17:58
Let $P=AE\cap (ABCD)$
$\textbf{Claim:}$ $M$ on $DP$
$\emph{Proof.}$ Let $M'=DP\cap EC$. Thus $\angle CEP=\angle AEB=\frac \pi 2 -\angle EAB=\frac \pi 2 -\angle DAC=\angle ACD=\angle APD$. So $\triangle M'PE$ is isosceles. Because $\angle APC=\angle ADC$, $\triangle EPC$ is right-angled. Therefore $M'$ is the midpoint of $EC$. So $M=M'$, $P$ on $DM$. $\Box $
So $\angle BDP=\angle BAP=\angle BAE=\angle DAC=\angle DBC$. Thus $BM=DM$ as desired. $\Box $
Theavenger1945
12.02.2022 18:05
songhongyi (Profile) $\cdot$ 6 minutes ago $\cdot$ Y 0 $\cdot$
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SolutionLet $P=AE\cap (ABCD)$
$\textbf{Claim:}$ $M$ on $DP$
$\emph{Proof.}$ Let $M'=DP\cap EC$. Thus $\angle CEP=\angle AEB=\frac \pi 2 -\angle EAB=\frac \pi 2 -\angle DAC=\angle ACD=\angle APB$. So $\triangle M'PE$ is isosceles. Because $\angle APC=\angle ADC$, $\triangle EPC$ is right-angled. Therefore $M'$ is the midpoint of $EC$. So $M=M'$, $P$ on $DM$. $\Box $
So $\angle BDP=\angle BAP=\angle BAE=\angle DAC=\angle DBC$. Thus $BM=DM$ as desired. $\Box $
lazizbek42
12.02.2022 18:50
$AE$ and $(ABCD)$ intersection $X$.$XD$ and $BC$ intersection $Y$.Very easy $Y$ is midpoint of $CE$
Mahdi_Mashayekhi
31.03.2022 09:37
Let $C'$ be reflection of $C$ across $D$, Let $AE$ meet $BD$ at $S$. $\angle SAB = \angle SBE \implies EB$ is tangent to $ASB$ which implies $\angle ASB = \angle 180 - \angle ABE = \angle ADC \implies \angle DAS = \angle SDC$. Note that for proving $BM = DM$ we need to prove $\angle MDB = \angle DAC$ or $\angle CAE = \angle CDM = \angle CC'E$ or $CC'AE$ is cyclic. Note that $ACC'$ is isosceles so $\angle AC'C = \angle ADC - \angle DAC' = \angle ADC - \angle DAC = \angle ASB - \angle SBE = \angle AEB$ so $CC'AE$ is cyclic as wanted. we're Done.