Plane Geometry related to Circle

Question

The internal bisector of angle A of triangle ABC meets the circumcircle in D. If DE and DF are the perpendiculars to AB and AC respectively from D. Prove that AE is arithematic mean of AB and AC .

Answer 1

Of course no trigonometry is required. Considering the Figure below, we need to prove that $\triangle CDF \cong \triangle BDE$. This is straightforward by noting the following.

1. $DE \cong DF$, since $AD$ is bisector of $\angle CBD$.
2. $\angle DFA \cong \angle DEB$, being both right angles, by construction.
3. $\angle DBE \cong \angle DCF$, since they're both supplementary w.r.t $\angle ACD$ (can you tell why?)

The thesis follows from the fact that $CF \cong EB$. Can you tell why?

Answer 2

Let $ABCD$ be as in the diagram below.

Then $AB=2R\cos\theta$ and $AC=2R\cos\phi$.

So $$\frac{AB+AC}{2}=R\cos\theta+R\cos\phi=2R\cos(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})=AD\cos(\frac{\theta-\phi}{2})$$ which is the perpendicular distance. (Note that angle $DAC=(\theta-\phi)/2$.)