(Iran TST 2009 P9) In triangle $ABC$, $D$, $E$ and $F$ are the points of tangency of incircle with the center of $I$ to $BC$, $CA$ and $AB$ respectively. Let $M$ be the foot of the perpendicular from $D$ to $EF$. $P$ is on $DM$ such that $DP = MP$. If $H$ is the orthocenter of $BIC$, prove that $PH$ bisects $ EF$.
Let $N$ be the midpoint of $EF$ we wish to show that $P, N$, and $H$ are collinear. We invoke a few lemmas.
Returning to the problem at hand, denote the feet of the altitudes in triangle $BIC$ from $B$ and $C$ in by $X$ and $Y$, respectively. By Lemma 2, we know immediately that $X$ and $Y$ lie on the line $EF$. Then there are two key claims.
Claim 1: The incircle of triangle $DXY$ is tangent to side $XY$ at point $N$.
Proof: First, we prove that $I$ is the incentre of triangle $DXY$. To do this, we'll show that $IX$ and $IY$ bisect $\angle{DXY}$ and $\angle{DYX}$, respectively. Since $E, X, I, C$ are concyclic,
\[\angle{YXI} = \angle{EXI} = \angle{ECI} = \frac{1}{2} \angle{C} \]On the other hand,
\[ \angle{IXD} = 90^{\circ} - \angle{DXC} = 90^{\circ} - \angle{DIC} = \frac{1}{2} \angle{C} .\]Hence $IX$ bisects $\angle{DXY}$. The proof for $IY$ is analgous. Thus $I$ lies on the intersection of the angle bisectors and must be in the incentre.
Importantly, we also have that $N$ is the point of tangency of in the incircle of triangle $DXY$ with side $XY$. This follows from the fact that $INF$ and $INE$ are congruent so $IN$ is perpendicular to $EF$. Next, we prove the second claim.
Claim 2: Point $H$ is the $D$-excentre of triangle $DXY$.
Proof: It is sufficient to show that $H$ lies on the external bisectors of $\angle{DXY}$ and $\angle{DYX}$. This is a simple angle chase. Note that
\[ \angle{HXY}= 90^{\circ} - \angle{YXI} = 90^{\circ} - \frac{1}{2} \angle{C} .\]Likewise,
\[ 180^{\circ} - \angle{DXH} = 180 - \angle{HXI} - \angle{IXD} = 180^{\circ} -\left(90^{\circ} + \frac{1}{2} \angle{C}\right) = 90^{\circ} - \frac{1}{2} \angle{C} = \angle{HXY} . \]The other side is, again, analgous.
Hence $H$ is the $D$-excentre of triangle $DXY$ and $N$ is the point of tangency of the incircle to $XY$. Since $P$ is the midpoint of the $D$-altitude, we have by Lemma 1, that the three points are collinear.
Lemma 2: Let $ABC$ be an arbitrary triangle with incentre $I$ and intouch triangle $DEF$. Ray $BI$ meets $EF$ at $K$. Then $BK$ is perpendicular to $CK$.
Proof
We angle chase to show that $\angle{IKC}=90^{\circ}.$ It suffices to prove that $K$ is on the circumcircle of cyclic quadrilateral $DCEI$. Indeed, bearing in mind that $K$ is on the $\angle{B}$ internal bisector and thus equidistant from $F$ and $D$,
\[ \angle{KDC} = 180^{\circ} - \angle{BDK} = 180^{\circ} - \angle{BFK} = \angle{AFE} = \angle{AEF} = \angle{KEC} . \]Hence $K$ lies on the circumcircle of points, $C, D, E$. This proves the lemma.
Lemma 1: Let $ABC$ be an arbitrary triangle. Let $AD$ be an altitude with the midpoint denoted $M$. Let $K$ be the point of tangency of the incircle to side $BC$ and $I_A$ be the $A$-excentre. Then the points $M, K, I_A$ are collinear. Proof
Let $EF$ be a diameter of the $A$-excircle parallel to $AD$, with point $E$ tangent to side $BC$. First, consider the homothety sending the incircle to the $A$-excircle. It sends $K$ to $F$ so $A, K, F$ are collinear. Because $AD$ and $FE$ are parallel, there exists a homothety sending $A$ to $F$ and $D$ to $E$. It is centred at the intersection of $AF$ and $DE$, which is simply $K$. However, this homothety also sends $M$ to $I_A$ since they are the midpoints of respective segments $AD$ and $FE$. Hence $M, K, I_A$ are collinear, as desired.
Let $A B C$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{B C}, \overline{C A}$, $\overline{A B}$ at $D, E, F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{E F}$. Suppose that the circumcircle of $\triangle A I B$ meets the incircle at two distinct points $C_{1}$ and $C_{2}$, while the circumcircle of $\triangle A I C$ meets the incircle at two distinct points $B_{1}$ and $B_{2}$. Prove that the radical axis of the circumcircles of $\triangle B B_{1} B_{2}$ and $\triangle CC_{1} C_{2}$ passes through the midpoint $M$ of $\overline{D K}$.
Proof. First let's do some labeling. Let $X, Y, Z$ be the midpoints of $\overline{E F}, \overline{F D}, \overline{D E}$, and let the circumcircles of triangles $B B_{1} B_{2}$ and $C C_{1} C_{2}$ be $\omega_{B}$ and $\omega_{C}$. Also, if $H$ is the orthocenter of $\triangle B I C$, let $\omega_{B}$ intersect $\overline{H B}, \overline{A B}, \overline{B C}$ at $R, R_{1}, R_{2}$. Similarly, let $\omega_{C}$ intersect $\overline{H C}, \overline{A C}, \overline{B C}$ at $S, S_{1}, S_{2}$.Observe that $X$ is the radical center of $(A E F),(D E F),(A I B),(A I C)$, so $X$ is in fact the intersection of $\overline{B_{1} B_{2}}$ and $\overline{C_{1} C_{2}}$. We can obtain similar results for $Y$ and $Z$; thus, $\overline{X Z}$ and $\overline{X Y}$ are just $\overline{B_{1} B_{2}}$ and $\overline{C_{1} C_{2}}$.
We must have $B_{1}, B_{2}$ symmetric about line $B I$, so $\omega_{B}$ is also symmetric about line $B I$. So $R_{1}$ and $R_{2}$ are reflections across $\overline{B I}$, so $\overline{R_{1} R_{2}} \| \overline{D F}$. Similarly, $\overline{S_{1} S_{2}} \| \overline{D E}$.
The key is the following claim. Claim. $R, R_{1}, S, S_{1}$ are collinear. Proof. To prove this, we do some angle chasing. We know that $\angle R_{2} R_{1} R=\angle R_{2} B R=$ $\angle D F E$ and $\angle S_{2} S_{1} S=\angle 180^{\circ}-\angle S_{2} C S=180^{\circ}-\angle D E F$. Finally, we have $\angle\left(R_{1} R_{2}, S_{1} S_{2}\right)=$ $\angle E D F$ from the parallel lines, so
$$
\angle R_{2} R_{1} R+180^{\circ}-\angle S_{2} S_{1} S+\angle\left(R_{1} R_{2}, S_{1} S_{2}\right)=180^{\circ}
$$
Thus $\overline{R_{1} R}$ and $\overline{S_{1} S}$ must be the same line, so all of them are collinear.
Advanced Lemmas in Geometry, Fedir Yudin中使用了以下例题:
Let the incircle $\omega$ of $\triangle A B C$ touch $A C$ and $A B$ at points $E$ and $F$ respectively. Points $X, Y$ of $\omega$ are such that $\angle B X C=\angle B Y C=90^{\circ}$. Prove that $E F$ and $X Y$ meet on the medial line of $A B C$.
Solution. Let $M_{a}, M_{b}, M_{c}$ be the midpoints of $B C, C A, A B$ and let $T=$ $M_{b} M_{c} \cap E F$. It suffices to prove that $T$ lies on the radical axis of $\omega$ and the circle with diameter $B C$. By Iran lemma, $E F$ and the circle with diameter $B C$ intersect at two points $P$ and $Q$, lying on $M_{a} M_{b}$ and $M_{a} M_{c}$, respectively. Then $M_{b} E \| M_{c} Q$ and $M_{c} F \| M_{b} P$, so $\frac{T E}{T Q}=\frac{T M_{b}}{T M_{c}}=\frac{T P}{T F} \Longrightarrow T E \cdot T F=T P \cdot T Q$ and the conclusion follows.
