Triangle $ABC$ is given, where $AC<BC$ and $\angle ACB=60^\circ\!\!.$ Point $D$, distinct from $A$, lies on the segment $AC$ such that $AB=BD$, and point $E$, distinct from $B$, lies on the line $BC$ such that $AB=AE$. Prove that $\angle DEC=30^\circ$.
Problem
Source: 2023 Polish Junior Mathematical Olympiad Finals
Tags: geometry
Tsikaloudakis
06.05.2023 14:00
στις υπόψιν προς λύση.
Tintarn
06.05.2023 18:42
In other words, we need to prove that $CE=DE$.
If $A$ and $C$ are fixed and $B$ moves linearly, then both $C$ and $E$ move linearly. So it suffices to check two cases.
If $ABC$ is equilateral, then $C=D=E$. If $B=C$, the claim is also trivial. Done.
By trigonomety: We have
\[\frac{CD}{\sin \angle CBD}=\frac{BD}{\sin \angle DCB}=\frac{AD}{\sin \angle DCB}=\frac{AE}{\sin \angle ECD}=\frac{CE}{\sin \angle CAE}\]and the claim follows since $\angle CBD=\angle CAE$.
sunken rock
15.05.2023 19:01
See the attached drawing: $m(\widehat{CAE})=60^\circ-m(\widehat{ABC}), m(\widehat{BAC})=120^\circ-m(\widehat{ABC})$, so $m(\widehat{BAE})=180^\circ-2m(\widehat{ABC})=2m(\widehat{BAC})-60^\circ$. Construct the equilateral triangle $\triangle ABK$, thus $m(\widehat{KAE})=2m(\widehat{BAC})$, meaning $m(\widehat{AEK})=90^\circ-m(\widehat{BAC})=m(\widehat{ABF}), F$ being the projection of $B$ onto $AC$, or $m(\widehat{ABD}=2m(\widehat{AEK})$, and if $O$ is reflection of $B$ over $AC, O$ is circumcenter of $\triangle AED'$, supposing $EK\cap AC=D'$, thus $D'\equiv D$ and $m(\widehat{CED})=m(\widehat{CEK})=30^\circ$, since $A$ is the circumcenter of $\triangle BEK$, done. Best regards, sunken rock
Attachments:
