Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National Olympiad Prove it Iran

In the triangle $ABC$ , variable points $D$ , $E$ , $F$ are on the sides $BC$ , $CA$ , $AB$ respectively such that the triangle $DFE$ is similar to the triangle $ABC$ in the same order as written. Circumcircles of $BDF$ and $CDE$ intersect the circumcircle of $ABC$ at $P$ and $Q$ , respectively for the second time. Prove that the circumcircle of $DPQ$ passes through a fixed point.

Solution

Let the tangent of $\odot(ABC)$ at $A$ intersect $BC$ at $G$ and $FE \cap AG = H$ , $FD \cap AC = I$ , $ED \cap AB = J$ . We'll show that $\odot(DPQ)$ passes through the fixed point $G$ .

We claim that $A$ , $H$ , $J$ , $Q$ , $E$ are concyclic. Analogously, $A$ , $F$ , $P$ , $H$ , $I$ are also concyclic. It follows from the angle chasing:
$\angle FED = \angle ACB = \angle HAB$
So $A$ , $H$ , $J$ , $E$ are concyclic. Also:
$\angle AQE = \angle AQC - \angle EQC = \angle ABC - \angle EDC = \angle AJE,$
Which means $A$ , $J$ , $Q$ , $E$ are also concyclic. These two cyclic quadrilaterals prove our claim. Notice that:
$\angle QDC = \angle QEC = \angle QHA$
So $G$ , $H$ , $D$ , $Q$ are concyclic. Likewise, $G$ , $H$ , $D$ , $P$ are concyclic. Therefore $G$ , $P$ , $D$ , $Q$ , $H$ are concyclic, which implies that $G$ lies on $\odot(DPQ)$ .

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