Incircle \omega of triangle \triangle ABC is centered at I and touches sides BC, CA, AB at D, E, F respectively. Let Q be a point on the circumference of \omega such that \angle AQD=90^o, and let M be the midpoint of side BC. Suppose P is a point inside \triangle ABC such that P is on AI and MD=MP. Prove that either \angle EQP=90^o or \angle FQP=90^o.
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Assume WLOG AB<AC.
Claim: D,P,E are collinear.
Proof: Let N be the midpoint of AB.
It is well-known that MN,DE,AI are concurrent at a point. Let P’ be this intersection point, noting that P’ lies on segment DE. Then P’ lies inside △ABC.∠DMP’=∠BMN=180-∠B-∠BNM,
Since MN∥AC,∠BNM=∠A
<DMP’=180-∠B-∠A=∠C.
∴ △DMP’ ~ △DCE (A.A)
Since EC=ED,MP’=MD. Hence P’= P which finishes the claim.
Let S be the point diametrically opposite D on the incircle, which is also the second intersection of AQ with the incircle. Let T= AQ ∩ BC.
Then T is the contact point of the A-excircle; consequently,MD=MP=MT and we obtain circle with diameter DT. Since ∠DQT=∠DQS=90° we have Q on this circle as well.
As SD is tangent to the circle with diameter DT, we obtain
∠PQD=∠SDP=∠SDE=∠SQE.
Since ∠DQS=90°,∠PQE=90° too.
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