Given $\frac{1}{AH^{2}}=\frac{1}{AB^{2}}+\frac{1}{AC^2}$, prove $AH$ is perpendicular to $BC$

Question

In the right triangle $\triangle$ABC at A with H $\in$ BC that: $$\frac{1}{AH^{2}}=\frac{1}{AB^{2}}+\frac{1}{AC^{2}},$$ prove that AH $\perp$ BC.

My idea is to draw AH' $\perp$ BC (H' $\in$ BC) and prove H $\equiv$ H'

$$AH' \perp BC \implies AH'.BC = AB.AC = 2.Area \triangle ABC\implies AH' = \frac{AB.AC}{BC}$$

$$\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{AB^2+AC^2}{AB^2.AC^2}=\frac{BC^2}{(AB.AC)^2}\implies AH=\frac{AB.AC}{BC}$$

Then AH = AH'

I think that if AH = AH' and H, H' both lies in BC, H and H' must be the same point and therefore AH $\perp$ BC (because AH' $\perp$ BC)

But how do I prove this part? I think I'm nearly there but a little stuck came in my way. All help is appreciated. Thank you.

Answer 1

You're essentially done. Although in general from $AH=AH'$ and $H,H'\in BC$ we can't conclude that $H=H'$, here it is true because $AH'$ is perpendicular to $BC$. The reason: $AH'$ is the shortest distance from $A$ to the line $BC$, and there's a unique point on $BC$, viz. the base of the perpendicular dropped from $A$, that yields the shortest distance.

To justify this even more convincingly, although that would probably be an overkill, consider the circle of radius $AH'$ centered at $A$. Since $AH'\perp BC$, the circle is tangent to $BC$, and thus intersects it only at one point.

Answer 2

Let $H $ be some point on $BC $ such that $(AC \,.\, AB) = (AH \,.\, BC).$ (One way of writing of the given condition). Divide both sides by $2$. RHS will be area of $ABC$. So $AH $ has to be perpendicular to $ BC$.