The ratio of the area of $\triangle ABC$ to the length of $EF$?

Question

In $\triangle ABC$, D is the foot of perpendicular from A on BC. If E and F are the feet of perpendiculars from D on AB and AC respectively, find the ratio of the area of $\triangle ABC$ to the length of EF.

Answer 1

It is easy to see that:

• $DE=h\cos B$
• $DF=h\cos C$
• $\angle EDF= \pi - \angle A$
• $\angle DEF= \frac{\pi}2 - \angle C$ $\text{ } $ (as $ADEF$ is cyclic)

Where $h=AD$. Hence by the sine rule:

$$EF=\frac{DF\sin A}{\cos C}=h\sin A=\frac{ah}{2R}=\frac{\Delta}{R}$$