Let $\omega$ be a circle and let $A,B,C,D,E$ be five points on $\omega$ in this order. Define $F=BC\cap DE$, such that the points $F$ and $A$ are on opposite sides, with regard to the line $BE$ and the line $AE$ is tangent to the circumcircle of the triangle $BFE$. a) Prove that the lines $AC$ and $DE$ are parallel b) Prove that $AE=CD$
Problem
Source: Italy National Olympiad 2020 P1
Tags: geometry, circumcircle
A.L.E.X
01.10.2020 00:12
a). $\angle BFE=\angle AEB=\angle ACB \Rightarrow AC\parallel DE$ b). Since $AEDC$ is cyclic and $AC\parallel ED\implies AE=CD$ $\blacksquare$
Mahdi_Mashayekhi
08.01.2022 17:34
a) ∠BCA = ∠BEA = ∠BFE ---> ED || AC b) ED || AC ---> AE = CD
SilverBlaze_SY
13.04.2024 22:28
Let $X$ be a point on line $AE$, on the opposite side of $A$ with respect to $E$. $ABCDE$ is a cyclic pentagon. $AE$ is tangent to the circumcircle of $BFE$. Therefore, by alternate segment theorem, we get that $\angle AEB=\angle BFE$. Again, as $AECB$ is cyclic, $\angle AEB=\angle ACB$. Therefore, as $\angle ACB=\angle EFB\implies ED\parallel AC$. Again, by alternate segment theorem, $\angle EBF=\angle FEX$. As $AEDC$ is cyclic, we have $\angle ACD=\angle FEX$ $\implies \angle EBF=\angle ACD=\angle ABD$ (same segment $AD$) $\implies \angle CBD+\angle EBD=\angle ABE+\angle EBD \implies \angle CBD=\angle ABE\implies AE=CD$