Let $ABCD$ be a cyclic quadrilateral inscribed in circle $O$, where $AC\perp BD$. $M$ be the midpoint of arc $ADC$. Circle $(DOM)$ intersect $DA,DC$ at $E,F$. Prove that $BE=BF$.
Problem
Source: CSMO Grade 11 Problem 6
Tags: geometry
githjijjj
06.08.2017 09:12
A rough sketch of the solution:
Firstly, by Spiral Similarity, $\triangle MEF \sim \triangle MAC$ but $MA=MC$ so $ME=MF$. Angle chasing also gives $\angle EMA=\angle CMF=\angle MCF=\angle MBD$. Let these angles be $x^\circ$. Consider a clockwise rotation centered at $M$. It sends the line $MC$ to $MF$, line $MA$ to $ME$, and $OM$ to some random line $l$. Let $l$ intersect $BD$ at $B'$. Since $OM\parallel B'D$, $\angle OMB'=x^\circ=\angle MB'D=\angle MBD$, so $B'=B$ and $l$ is in fact $MB$. Combined with the fact that $\triangle MEF \sim \triangle MAC$ and $OM\perp AC$, we have $BM\perp EF$ and then we are done.
CJA
06.08.2017 09:35
l1090107005 wrote: Let $ABCD$ be a cyclic quadrilateral inscribed in circle $O$, where $AC\perp BD$. $M$ be the midpoint of arc $ADC$. Circle $(DOM)$ intersect $DA,DC$ at $E,F$. Prove that $BE=BF$. https://artofproblemsolving.com/community/q1h1487737p8738745
bobaboby1
22.08.2017 18:44
I think this problem is really boring.
China_BW
11.07.2022 11:48
bobaboby1 wrote: I think this problem is really boring. I think so~
f6700417
12.08.2022 14:13
Obviously, ΔMEA≌ΔMFC, ∴ME=MF. We only need to prove that the center of circle(MOD)∈MB. MB∩circle(MOD)=K. ∵∠MKO=∠MDO=90°-∠ODN=90°-∠MND=90°-∠NMB=90°-∠OMK, ∴∠MOK=90°.