The diagonals of a cyclic quadrilateral $ABCD$ meet at point $P$. The bisector of angle $ABD$ meets $AC$ at point $E$, and the bisector of angle $ACD$ meets $BD$ at point $F$. Prove that the lines $AF$ and $DE$ meet on the median of triangle $APD$.
Problem
Source: 10.1 of XX Geometrical Olympiad in honour of I.F.Sharygin
Tags: geo, geometry
NO_SQUARES
07.08.2024 09:49
Note that $\Delta ABP \sim \Delta DCP$ and so \[ \frac{AE}{EP}=\frac{AB}{BP}=\frac{DC}{CP}=\frac{DF}{FP} \]and thus problem is solved by Cheva's theorem for triangle $APD$.
ehuseyinyigit
07.08.2024 11:01
Let $AF\cap DE=Q$ and $PQ\cap AD=G$. Since $\dfrac{CD}{AB}=\dfrac{CP}{PB}$, we have by Ceva's Theorem
$$\dfrac{DF}{FP}\cdot \dfrac{PE}{EA}=\dfrac{CP}{PB}\cdot \dfrac{PB}{CP}=1=\dfrac{GD}{AD}$$which means $PG$ is the median of $\triangle APD$ and point $Q$ lies on it.
Couldn't participated Sharygin Geometry Olympiad this year but I am planning to compete for it in 2025. Suppose that I made the final exam, where will I take the exam as a foreign contestant ?
nervy
07.08.2024 11:40
ehuseyinyigit wrote:
Let $AF\cap DE=Q$ and $PQ\cap AD=G$. Since $\dfrac{CD}{AB}=\dfrac{CP}{PB}$, we have by Ceva's Theorem
$$\dfrac{DF}{FP}\cdot \dfrac{PE}{EA}=\dfrac{CP}{PB}\cdot \dfrac{PB}{CP}=1=\dfrac{GD}{AD}$$which means $PG$ is the median of $\triangle APD$ and point $Q$ lies on it.
Couldn't participated Sharygin Geometry Olympiad this year but I am planning to compete for it in 2025. Suppose that I made the final exam, where will I take the exam as a foreign contestant ?
Try to ask someone in your country to write a letter to the organizer about this. The organizers usually allow the mirror of the Olympiad to be held and throw off all materials for this.
ehuseyinyigit
07.08.2024 11:49
I got it. Surely teachers from our olympiad committe will help. Thanks