The circle inscribed in triangle $ABC$ touches $AC$ at point $F$. The perpendicular from point $F$ on $BC$ intersects the bisector of angle $C$ at point $N$. Prove that segment $FN$ is equal to the radius of the circle inscribed in triangle $ABC$.
(Oleksii Karliuchenko)
parmenides51 wrote:
The circle inscribed in triangle $ABC$ touches $AC$ at point $F$. The perpendicular from point $P$ on $BC$ intersects the bisector of angle $C$ at point $N$. Prove that segment $PN$ is equal to the radius of the circle inscribed in triangle $ABC$.
(Oleksii Karliuchenko)
I think the touch-point should be $P$. In that case let $H$ be the foot of perpendicular from $P$ on $BC$. We have $\angle HNC = 90-\frac{C}{2}$ and $\angle CIP = 90-\frac{C}{2}$ which yields $PIN$ is isosceles so $PN=IP$.