Understand the problem

Let $ABC$ be a triangle and $AD,BE,CF$ be cevians concurrent at a point $P$. Suppose each of the quadrilaterals $PDCE,PEAF$ and $PFBD$ has both circumcircle and incircle. Prove that $ABC$ is equilateral and $P$ coincides with the center of the triangle.
Source of the problem
Indian team selection test 2018
Topic
Geometry
Difficulty Level
Hard

Suggested Book
Challenge and Thrill of Pre-college Mathematics by B J Venkatachala, C R Pranesachar, K N Ranganathan and V Krishnamurthy

Start with hints

Do you really need a hint? Try it first!

Quadrilaterals having both an incircle and a circumcircle are called bicentric. Read more about them here and here.
Show that BCEF is cyclic. Afterwards, study the consequences of this result.    

As \angle BFC+\angle BEC=\pi, hint 1 means that \angle BFC=\angle BEC =\frac{\pi}{2}. Hence P is the orthocentre of ABC. Let T be the circumcenter of AFPE and L be its incentre. From the information given in the second link of hint 1, we see that T,L,K are colinear.Also, as \angle AEP=\frac{\pi}{2}T lies on AP.

As L lies on APAD is the internal bisector of \angle A. As AD is also an altitude, this means that AB=AC. By symmetry, BC=AB and CA=BC. Hence the triangle is equilateral.

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