PH bisects EF !

Problem :

Let $ABC$ be a triangle with incenter $I$  and contact triangle $DEF$. Let $M$ be the foot of perpendicular from $D$ to $EF$ and let $P$ be the midpoint of $\overline{ \rm DM }$. If $H$ is orthocenter of triangle $BIC$, prove that $ \overline{ \rm PH } $ bisects $ \overline{ \rm EF } $.

( Source : Iran TST 2009)

Solution :

This is a really nice geometric configuration. Before going into the solution one must read Let's Look Into The Incircle as we are going use the properties discussed over there again and again to solve this problem. One might find the problem too easy if they remember all the lemmas discussed in Let's Look Into The Incircle, but otherwise this is really a very tough problem. So let's get started !

Observe the condition that $PH$ bisects $EF$ is equivalent to showing $ PH, EF $ and $ AI $ are concurrent ( as $AI$ bisects $EF$ ).

Let $ AI \cap EF = T $.

We will show that $ P, H $ and $T$ are collinear, which will complete the proof.

Let ray $BI$ meet $EF$ at $X$ and let ray $CI$ meet $EF$ at $Y$.

Now we have already proved in Let's Look Into The Incircle ( Lemma $1$ ) that,

• $ BX \perp CX $ and $ CY \perp BY $.
• $I$ is incenter of triangle $DXY$.

Again by Lemma $2$ in Let's Look Into The Incircle we get that,

$\triangle DXY $ is orthic triangle of $ \triangle HBC $, and $I$ is the orthocenter of $\triangle HBC $

Hence $H$ is the $D$-excenter of $ \triangle DXY $.

Thus, simplifying the problem statement we have to prove that in triangle $DXY$, $P := $ the midpoint of altitude from $D$ to $XY$,  $T := $ the intouch point of incircle of $\triangle DXY $ on $XY$ and the $H := D$-excenter of $\triangle DXY $ are collinear, which we have already proved in Let's Look Into The Incircle ( Lemma $4$ ). Hence $ P, T $ and $H$ are collinear.

Hence, proved $PH$ bisects $EF$.