Altitudes of Triangle Meet at Point/Proof 2

Theorem

Let $\triangle ABC$ be a triangle.

The altitudes of $\triangle ABC$ all intersect at the same point.

In $\triangle ABC$ construct the altitude from vertex $A$ to side $BC$ at $P$.

Also draw the altitude from vertex $B$ to side $AC$ at $Q$.

$\angle AOQ = \angle BOP$

Given:

$\angle AQO = \angle BPO = $ one right angle.

$\triangle AOQ \sim \triangle BOP$

and:

$\triangle AOQ \sim \triangle APC \sim \triangle BOP$

By the definition of similarity:

$\dfrac {OP} {BP} = \dfrac {CP} {AP}$

Rearrange:

$OP = \dfrac {BP \cdot CP} {AP}$

Note there is no term related to side $AC$.

If we had used side $AB$ instead of $AC$, we would have the same result.

Therefore, the altitude from vertex $C$ to side $AB$ meets both $AP$ and $BQ$ at $O$.

$\blacksquare$