MathLabs

Problem 6

Let II be the incenter of acute triangle ABCABC with AB≠ACAB\ne AC. The incircle ω\omega of ABCABC is tangent to BCBC, CACA, and ABAB at DD, EE, and FF, respectively. The line through DD perpendicular to EFEF meets ω\omega again at RR (other than DD). Line ARAR meets ω\omega again at PP (other than RR). The circumcircles of triangles PCEPCE and PBFPBF meet again at QQ (other than PP). Prove that lines DIDI and PQPQ meet on the line through AA perpendicular to AIAI.
Step 6 of 6: Return to the original statement
Previous stepNext step
In plain words

The two circles through P,M,IP,M,I coincide.

P,M,I,Q,T are concyclic⟹T∈PQP,M,I,Q,T\text{ are concyclic}\Longrightarrow T\in PQ
Detailed analysis

The circles through P,M,I,TP,M,I,T and through P,M,I,QP,M,I,Q are the same circle, so P,Q,TP,Q,T are collinear in the inverted configuration. Undoing the inversion converts this collinearity into the original incidence T∈PQT\in PQ. Since TT was defined on DIDI and on the line through AA perpendicular to AIAI, the required conclusion follows.