The quadrilateral $ABCD$ is inscribed in the circle ω. The diagonals $AC$ and $BD$ intersect at the point $O$. On the segments $AO$ and $DO$, the points $E$ and $F$ are chosen, respectively. The straight line $EF$ intersects ω at the points $E_1$ and $F_1$. The circumscribed circles of the triangles $ADE$ and $BCF$ intersect the segment $EF$ at the points $E_2$ and $F_2$ respectively (assume that all the points $E, F, E_1, F_1, E_2$ and $F_2$ are different). Prove that $E_1E_2 = F_1F_2$. $(N. Sedrakyan)$
Problem
Source: Silk Road Mathematical Competition 2017, P2
Tags: geometry proposed, geometry
26.05.2017 06:56
any solution
28.05.2017 11:51
10.12.2022 11:40
The quadrilateral $ABCD$ is inscribed in the circle $\omega$. The diagonals $AC$ and $BD$ intersect at the point $O$. On the segments $AO$ and $DO$, the points $E$ and $F$ are chosen, respectively. The straight line $EF$ intersects $\omega$ at the points $E_1$ and $F_1$. The circumscribed circles of the triangles $ADE$ and $BCF$ intersect the segment $EF$ at the points $E_2$ and $F_2$ respectively (assume that all the points $E, F, E_1, F_1, E_2$ and $F_2$ are different). Prove that $E_1E_2 = F_1F_2$. $(N. Sedrakyan)$
26.07.2024 12:24
Extend $DE_2, CF_2$ to intersect $\omega$ again at $X,Y$ respectively. We have $\angle CXD = \angle CAD = \angle XE_2E \Rightarrow CX\parallel EF$. Similarly we can show $DY \parallel EF$. Hence $CX \parallel DY \parallel EF$, and thus $CXYD$ is an isosceles trapezium. Then $E_2F_2YD$ is also an isosceles trapezium by symmetry, and since $E_1F_1DY$ is also an isosceles trapezium, $E_1E_2 = F_1F_2$.