The internal and external bisectors of $\angle A$ meet line $BC$ at $E$ and $F$. Show that the tangent at $A$ to $\bigcirc ABC$ bisects $EF$. 
Suppose the internal and external bisector of $\angle A$ meet the side $BC$ (produced) at $E$ and $F$, respectively. If the tangent at $A$ to $\bigcirc ABC$ meets $BC$ (produced) at $D$, prove that $D$ bisects $EF$.

 A: Hint: Simple angle chasing.
Note that $AEF$ is a right triangle, so to show that $D$ is the midpoint, it suffices to thow that $ADE$ is isosceles with $DA = DE$.   

 We have $ \angle DEA = \frac{\alpha }{2} + \beta$ by exterior angle of triangle and $\angle DAE = \beta + \frac{\alpha}{2}$ by alternate segment theorem.    

A: 
Note $\angle DAB = \angle ACD = \alpha $ due to tangent line AD. Then, $\angle DAE = \angle DEA = \alpha + \angle BAE$ because AE bisects $\angle BAC$. The triangle ADE is isosceles.
Also note that AE $\perp$ AF due to the angle bisectors AD and AE. Then, the triangle AFD is isosceles because of the isosceles triangle ADE. Thus, DE = DA = DF and D is the midpoint.
A: Let $c>b$. Let $K$ be a midpoint of $EF$. We have $\frac{EC}{EB}=\frac{b}{c}$, $\frac{FC}{FB}=\frac{b}{c}$, $EC+EB=a$, $FB-FC=a$. Then $EC=\frac{ab}{b+c}$, $EB=\frac{ac}{b+c}$, $FC=\frac{ab}{c-b}$. Then $EF=EC+FC=\frac{2abc}{c^2-b^2}$. Since $AK$ is median of triange $AEF$ with $\angle EAF=\frac{\pi}{2}$ then $AK=\frac{1}{2}EF$. Then $AK=\frac{abc}{c^2-b^2}$. Also we have $KC=KE-EC=\frac{ab^2}{c^2-b^2}$, $KB=KE+EB=\frac{ac^2}{c^2-b^2}$. Since $KC=\frac{ab^2}{c^2-b^2}$, $KB=\frac{ac^2}{c^2-b^2}$ and $AK=\frac{abc}{c^2-b^2}$ then easy to see that $KA^2=KC \cdot KB$. Since $KA^2=KC \cdot KB$ then $AK$ is tangent at $A$ to the circumcircle of triangle $ABC$. Then $K=D$. Then $D$ is midpoint of $EF$.
