<s>[INST] 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 $EF$.

 [/INST] 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 $EF$.
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