Suppose the internal and external bisector of $\angle A$ meet the side $BC$ and $BC$ produced at $E$ and $F$ respectively. If the tangent at $A$ to the circle $ABC$ meets $BC$ produced at $D$ , prove that $D$ bisects $EF$.
- I have tried various approaches to solve this problem.I extended the line $AD$ upto $A'$ such that $AD$ is equal to $DA'$. Then I wanted to show that $AF$ is parallel to $EA'$ which will prove that $FD = ED$. But this approach isn't working.
- Another approach I tried is ,I created a line $FA'$ equal and parallel to $AE$. Then I proved that $AEA'F$ is a parallelogram.Then I wanted to show that $ADA'$ is a straight line which will prove that $D$ bisects $EF$ , since diagonals of a parallelogram bisect each other. This approach isn't also working.
- I want this problem to be solved using basic theorem about circle and triangles in euclidean geometry.
Let $\measuredangle ABC<\measuredangle ACB$.
$$\measuredangle EAD=\measuredangle EAC+\measuredangle CAD=\measuredangle B+\measuredangle EAB=\measuredangle AED,$$ which says that $AD=ED$ and since $\measuredangle EAF=90^{\circ}$, we obtain $AD=DF$, $ED=DF$
and we are done!