chaotic_iak wrote: The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.

Hint

There's a reason why this is in Proposed & Own Problems (or otherwise known as "solved by poster, but looking for alternate solutions"). I didn't solve this in the test because I tried on the Algebra one, and when I see this, I immediately think of analytic geometry (hey, one circle, one bisector; use the bisector as an axis and we have a good fact that the gradients of the two sides of the bisected angle are opposites!)...which leads to something messy. So I gave up. Without thinking of simple angle chasing. :headbash: (Also, I get the one-liner solution probably precisely as you want to say from a friend.)

chaotic_iak wrote: The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $\color{green}\boxed{\textbf{Solution 1:}}$ $\color{green}\rule{24cm}{0.3pt}$ Let $I$ be the incenter of $\triangle ABC$ Let $\angle MBI=\angle IBC=\alpha, \angle NCI=\angle ICB=\beta$ $$\Rightarrow \angle ABC=2\alpha, \angle ACP=2\beta$$$$\Rightarrow \angle BAC=180-2\alpha -2\beta$$$$\Rightarrow \angle NAI=90-\alpha -\beta...(I)$$Since $\angle IMA=\angle INA=90$ $$\Rightarrow ANIM \text{ is cyclic}$$$$\Rightarrow \angle NMI=\angle NAI$$By $(I):$ $$\Rightarrow \angle NMI=90-\alpha-\beta$$In $\triangle BMP:$ $$\alpha +90+90-\alpha -\beta +\angle MPB=180$$$$\Rightarrow \angle MPB=\beta=\angle NCI$$$$\Rightarrow PNIC \text{ is cyclic}$$$$\Rightarrow \angle IPC=\angle INC=90$$$$\Rightarrow \boxed{BP\perp CP}_\blacksquare$$$\color{green}\rule{24cm}{0.3pt}$ $\color{green}\boxed{\textbf{Solution 2:}}$ $\color{green}\rule{24cm}{0.3pt}$ Let $T$ be the point of tangency of the incircle to the side $BC$ Working in complex numbers, where $\odot (MNP)$ is the unit circle, let $m=m$, $n=n$, $t=1$ and $i=0$ $A$ is the intersection of the tangents to the incircle that pass through $M$ and $N:$ $$\Rightarrow a=\frac{2mn}{m+n}$$Analogously we have to: $$b=\frac{2m}{m+1}$$$$c=\frac{2n}{n+1}$$$P, B \text{ and }O$ are collinears: $$\Rightarrow \frac{p}{\overline{p}}=\frac{b}{\overline{b}}$$$$\Rightarrow \frac{p}{\overline{p}}=\frac{\frac{2m}{m+1}}{\frac{2}{m+1}}=m$$$$\Rightarrow \overline{p}=\frac{p}{m}...(\blacktriangle)$$$M, N \text{ and }P$ are collinears: $$\Rightarrow \frac{p-m}{\overline{p}-\overline{m}}=\frac{m-n}{\overline{m}-\overline{n}}=-mn$$$$\Rightarrow \overline{p}=\frac{m+n-p}{mn}...(\blacksquare)$$By $(\blacktriangle)$ and $(\blacksquare):$ $$\Rightarrow \frac{p}{m}=\frac{m+n-p}{mn}$$$$\Rightarrow p=\frac{m+n}{n+1}...(\star)$$$BP\perp CP:$ $$\Leftrightarrow \frac{b-p}{\overline{b}-\overline{p}}=-\frac{c-p}{\overline{c}-\overline{p}}$$By $(\star):$ $$\Leftrightarrow \frac{\frac{2m}{m+1}-\frac{m+n}{n+1}}{\frac{2}{m+1}-\frac{m+n}{m(n+1)}}=-\frac{\frac{2n}{n+1}-\frac{m+n}{n+1}}{\frac{2}{n+1}-\frac{m+n}{m(n+1)}}$$$$\Leftrightarrow \frac{2m(n+1)-(m+n)(m+1)}{\frac{2m(n+1)-(m+n)(m+1)}{m}}=-\frac{n-m}{\frac{m-n}{m}}$$$$\Leftrightarrow m=m\checkmark$$$$\Rightarrow \boxed{BP\perp CP}_\blacksquare$$$\color{green}\rule{24cm}{0.3pt}$ $\color{blue}\rule{24cm}{0.3pt}$

It's just Iran Lemma