Given is an isosceles triangle $ABC$ with $CA=CB$ and angle bisector $BD$, $D \in AC$. The line through the center $O$ of $(ABC)$, perpendicular to $BD$, meets $BC$ at $E$. The line through $E$, parallel to $BD$, meets $AC$ at $F$. Prove that $CE=DF$.
Problem
Source: Serbia JBMO TST 2023 P1
Tags: geometry
Orestis_Lignos
28.05.2023 20:55
Note that $\dfrac{DF}{DC}=\dfrac{BE}{BC}$, and so it suffices to prove that $\dfrac{CE}{BE}=\dfrac{DC}{BC}$. Note that $\angle COE=\angle DBA=\angle DBC$ and $\angle EOB=\angle COB-\angle COE=2\angle CAB-\angle DBC=180^\circ-\angle C-\angle DBC=\angle CDB$. By the Ratio Lemma, $\dfrac{CE}{BE}=\dfrac{CO}{BO} \cdot \dfrac{\sin \angle COE}{\sin \angle EOB}=\dfrac{\sin \angle CBD}{\sin \angle CDB}=\dfrac{CD}{BC},$ and we may finish.
bin_sherlo
06.10.2023 18:09
Let $M$ be the midpoint of $AC$ and $OM \cap BD=K$. $BD \cap (ABC)=N$,$NE \cap AC=S$,$EO \cap AC=T$,$EO \cap NB=P$,$EO \cap AB=G$. Let $\angle CAB=\angle AB=2\alpha$ Claim: $NE \parallel AB$ Proof: $EO$ bisects $NB$ and $EO \perp NB \implies EB=EN \implies \angle ABN=\alpha=\angle NBE=\angle ENB$ So $NE \parallel AB$ as desired. Claim: $EK \parallel AC$ Proof: $\angle OPB=90\angle OMB \implies O,P,M,B$ are cyclic. $\angle KOP=\angle MOP=\angle MBP=\alpha=\angle KBE \implies O,K,E,B$ are cyclic. $\angle KEO=\angle KBO=\angle ABO-\angle ABK=4\alpha-90-\alpha=3\alpha-90$ $\angle TGA=\angle OGB=90-\alpha$ and $\angle CAB=2\alpha \implies \angle ATG=3\alpha-90$ $\angle ATE=3\alpha-90=\angle KET \implies AC \parallel EK$ as desired. We have $FE \parallel NB$ and $EK \parallel AC$ so $FDKE$ is a parallelogram. Also $\angle CKE=\angle OBC=90-2\alpha=\angle ECK \implies EC=EK$ We have $FD=EK=EC \implies FD=EC$.