Show that 3 circles related to a triangle intersects at common point

Question

We have triangle $ABC$ and points $D,E,F$ which lies repectively at $BC,AC,AB$. There are circles passing through $AFE$, $FDB$, $CDE$ show that they intersect at common point

Answer 1

Let $P$ be the "other" point of intersection of $\bigcirc AEF$ and $\bigcirc BFD$, and assume that $P$ is not exterior to $\triangle ABC$. (Minor tweaks to the argument handle the exterior case.) Since each of $\square AEPF$ and $\square BFPD$ is cyclic, opposite angles within them are supplementary. Now, observe,

$$\begin{align} \angle DPE &= 360^\circ - \angle EPF - \angle FPD \\ &= 360^\circ - (\; 180^\circ - \angle A \;) - (\; 180^\circ - \angle B \;) \\ &=\angle A + \angle B \\ &= 180^\circ - \angle C \end{align}$$

Having demonstrated that opposite angles of $\square CDPE$ are supplementary, we conclude that quadrilateral is also cyclic, which is to say: $P$ lies on $\bigcirc CDE$ (and therefore all three circles).