Solution: Let $D’$ be the reflection of $D$ upon $AI$. Clearly, $DD’||PQ$, $\angle DPQ=\angle D’QP$, therefore $DD’PQ$ is an isosceles trapezium which means $P,D,D’,E,Q$ all lie on a circle, which is absurd since obviously $DD’EQ$ is a concave quadrilateral. Thus $D’=E$. And so, $\angle DIA=\angle EIA\implies 45^{\circ}+\frac{\angle C}{4}=45^{\circ}+\frac{\angle B}{4}\implies \angle B=\angle C$. And we are done. $\blacksquare$

Are u really that imo medalist anant mudgal Posted a question on RMO group$ @ayan. math$

Paragdey12 wrote: Are u really that imo medalist anant mudgal Posted a question on RMO group$ @ayan. math$ Yes, he is.

ayan.nmath wrote: Solution: Let $D’$ be the reflection of $D$ upon $AI$. Clearly, $DD’||PQ$, $\angle DPQ=\angle D’QP$, therefore $DD’PQ$ is an isosceles trapezium which means $P,D,D’,E,Q$ all lie on a circle, which is absurd since obviously $DD’EQ$ is a concave quadrilateral. Thus $D’=E$. And so, $\angle DIA=\angle EIA\implies 45^{\circ}+\frac{\angle C}{4}=45^{\circ}+\frac{\angle B}{4}\implies \angle B=\angle C$. And we are done. $\blacksquare$ I didn't get this concave thing argument. Like, the configuration could also be something like below: [asy][asy] pair D=dir(-110),Dp=dir(-70),P=dir(-160),Q=dir(-20),E=dir(30),Ep=dir(150),B=extension(Ep,D,Dp,E); draw(unitcircle,magenta); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$E'$",Ep,dir(Ep)); dot("$D'$",Dp,dir(Dp)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$B$",B,dir(B)); draw(Ep--B--E,royalblue); [/asy][/asy] So someone please help.