show that three points are collinear

Question

Let $ABC$ be a triangle with $AB \neq AC$. Point $E$ is such that $AE = BE$ and $BE \perp BC$. Point $F$ is such that $AF = CF$ and $CF \perp BC$. Let $D$ be the point on line $BC$ such that $AD$ is tangent to the circumcircle of triangle $ABC$. Prove that $D,E,F$ are collinear.

Answer 1

Let $EK \perp AB$ at $K$ on $AB$, $AH \perp BC$ at $H$ on $BC$. $\triangle EBK \sim \triangle ABH \Rightarrow$ $$ \frac{EB}{AB}=\frac{BK}{AH}=\frac{AB}{2AH} $$ Similarly $$ \frac{FC}{AC}=\frac{AC}{2AH} $$ From which it follows: $$ \frac{EB}{FC}=\left(\frac{AB}{AC}\right)^2 $$ On the other hand, $\triangle ADB \sim \triangle ADC$, we have $$ \frac{DB}{AD}=\frac{AB}{AC}=\frac{AD}{DC}\Rightarrow \frac{DB}{DC}=\left(\frac{AB}{AC}\right)^2 $$ Therefore, $\frac{EB}{FC}=\frac{DB}{DC} \Rightarrow D,E,F $ are collinear.