Quote: In triangle $ABC$ with the incenter $I$ the bisectors of $\angle{B}$ , $\angle{C}$ intersect the sides $AC$ , $AB$ at $P$ , $Q$ respectively. Prove that $IP=IQ\ \iff\ AB=AC\ \vee\ A=60^{\circ}$ . Proof. $\left\|\begin{array}{c} m(\angle BPC)=x=A+\frac B2\\\\ m(\angle CQB)=y=A+\frac C2\end{array}\right\|\ \implies\ \frac cb=$ $\frac {[AIB]}{[AIC]}=$ $\frac {c\cdot IQ\cdot\sin y}{b\cdot IP\cdot\sin x}$ . $IP=IQ\iff\sin x=$ $\sin y\iff x=$ $y\ \vee\ x+y=\pi$ $\iff$ $b=c\ \vee\ A=60^{\circ}$ .

Its really easy~~ Consider the circumcircle of $ \triangle ABC $ . Since $ IP = IQ $, hence $ I $ lies in the perpendicular bisector of $ PQ $. Therefore we get $ A,P,I,Q $ are concyclic. Now from here we get $ \angle IAQ = \angle IPQ $ and the result follows. ($ \angle A = 60^{\circ} $ )

Poland NMO 2000 (half-final)

We have sin ∠IAP / sin ∠API = sin ∠IAQ / sin ∠AQI and IAP = IAQ so sin ∠API = sin ∠AQI. so we have to cases : 1- ∠API = ∠AQI : ∠API = ∠C + ∠B/2 and AQI = ∠B + ∠C/2 so ∠B = ∠C but we have AB > AC so Contradiction. 2 - ∠API + ∠AQI = 180 : AQIP is cyclic and ∠BIC = 90 + ∠A/2 ---> ∠A + 90 + ∠A/2 = 180 ---> ∠A = 60 we're Done.