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# 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
Geometry
Hard

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

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 $AP$$AD$ 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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