Reciprocal Altitude Theorem

Question

If $ABC$ is a triangle such that $D \in (BC) $ , $AD^2 =BD \cdot CD$ and $AD= \frac{AB \cdot AC}{BC} $ show that $ABC$ is a right triangle. I tried to solve it with Stewart Theorem but the calculus are not very frendly .

Answer 1

Notation: $AB=c$, $AC=b$, $BC=a$ , $AD=d ,CD=x,BD=a-x $ . We can write the second condition as :

$\frac{d}{c}= \frac{b}{a} \Rightarrow \frac{\sin B}{\sin \angle ADB} = \frac{\sin B}{\sin A } \Rightarrow \angle ADB = A $ or $\angle ADB= \pi-A$

That means that one of triangles ABD or ADC is similar to triangle ABC.

Wlog, let $\triangle ABD \sim \triangle ABC$ ($\angle ADB=\angle A)$

From there we get following ratio:

$ \frac{x}{c}=\frac{c}{a} \Rightarrow ax=c^2 $

$d^2=x(a-x)=ax-x^2=c^2-x^2 \Rightarrow x^2+d^2=c^2 \Rightarrow \angle ADB= 90^\circ = \angle A$

So the triangle ABC is right triangle with $A=90^\circ$