Let $ABCD$ be a convex quadrilateral, such that $AB = CD$, $\angle BCD = 2 \angle BAD$, $\angle ABC = 2 \angle ADC$ and $\angle BAD \neq \angle ADC$. Determine the measure of the angle between the diagonals $AC$ and $BD$.
Problem
Source: Rioplatense L-2 2023 #2
Tags: geometry, excenter, Angle Chasing
lambda5
06.12.2023 15:06
Kinda of nice geo problem (unexpected from RP )
It's natural to extend AB and CD so they encounter at $L$. The exincenter of $LBC$ from $L$ finishes the problem instantly after some trivial angle chase
SBYT
09.12.2023 11:44
Let $AB$ meets $CD$ at $L$,easy to know that $\angle BLC=\frac{\pi}{3}$.Let $\triangle BXC$ is a equilateral triangle($X,A$ lies on the same side of $BC$),we can know that $\triangle XBA\cong\triangle XCD$ and $B,C,L,X$ lie on the same circle($O$ is the center).$\angle XDL=\frac{\pi}{3}-\angle LXD=\frac{\pi}{3}-\angle LAD=\angle BAD-\frac{2\pi}{3}=2\angle BCD-\frac{2\pi}{3}=2(\angle LCB-\angle BCX)=2\angle LCX$,it means that $XD=DC$.We also have $XO=OC$,$XB=BC$,so $B,O,D$ are collinear,similarly we can get $A,O,C$ are collinear.Then $\measuredangle (AC,BD)=\pi-\angle BOC=\frac{\pi}{3}$.
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