Let $ABC$ be an isosceles triangle with $AB = AC$. A semi-circle of diameter $[EF] $ with $E, F \in [BC]$, is tangent to the sides $AB,AC$ in $M, N$ respectively and $AE$ intersects the semicircle at $P$. Prove that $PF$ passes through the midpoint of $[MN]$.
Problem
Source: JBMO Shortlist 2003
Tags: geometry, JBMO
sunken rock
15.10.2017 08:03
See that $AE$ is symmedian of $\triangle EMN$.
Best regards, sunken rock
wu2481632
15.10.2017 19:19
Suffices to show that $FP$ and $FA$ are isogonal in $\angle{MFN}$, which is trivially true.
stronto
05.01.2018 10:25
Different solution: let $D$ be the foot of the angle-bisector, then $D$ has to be the center of the circle by the tangency condition (equidistant from sides) -> implies that $AMDN$ is cyclic. Then, we also have $\angle APF = \angle EPF = 90$, but since $\triangle ABC$ is isosceles, we have $\angle ADF = 90$, thus, $APDF$ is cyclic. By radical axis, $MN$, $FP$, and $AD$ concur, but $AD$ is the perpendicular bisector of $MN$, Q.E.D