How to prove that AE is the arithmetic mean of AB and AC

Question

The internal Bisectors of angle A of triangle ABC meets the circumcircle at D. If DE,DF are the perpendiculars to AB , AC respectively from D,then prove that AE=(AB+AC)/2

Answer 1

Note that $BD=DC$ because inscribed angles $\angle BAD$ and $CAD$ are equal. Also $DE=DF$ (because point D lies on the bisector of $\angle A$) and $\angle DEB=\angle DFC$.

It follows by SSA that triangles $\triangle DEB$ and $\triangle DFC$ are congruent (usage od SSA is allowable because the angle in question is opposed to the longest side).

It means that $BE=FC=x$.

It is also obvious that triangles $\triangle ADE$ and $\triangle ADF$ are congruent by ASA ($AD=AD$ with all angles being equal: $\angle EAD=\angle FAD$, $\angle DEA=\angle DFA$, $\angle EDA=\angle FDA$). It follows that $AE=AF$.

The rest is trivial:

$$AB=AE+x$$

$$AC=AF-x=AE-x$$

$$AB+AC=2AE$$

Answer 2

• $\overline{DB}\cong\overline{DC}$, as chords subtending congruent inscribed angles.
• $\angle ABD$ and $\angle ACD$ are supplementary, as opposite angles in an inscribed quadrilateral.

Thus, we can extend $\overline{AB}$ by distance $|AC|$ to point $C^\prime$, so that $\triangle BC^\prime D\cong\triangle BCD$. Since $\triangle ADC^\prime$ is isosceles, altitude $\overline{DE}$ bisects the base, and the result follows. $\square$