Orthocenter

The orthocenter of a triangle is the point of intersection of its altitudes.

Proof that the altitudes of a triangle are concurrent

Using the trigonometric version of Ceva's Theorem it suffices to show that $\sin BAD\cdot \sin EBC\cdot \sin FCE = \sin CAD\cdot \sin ABE\cdot \sin BCF$ . Using the right angles gives us

$\sin BAD = \frac{HF}{AH}$
$\sin EBC = \frac{HD}{BH}$
$\sin FCE = \frac{HE}{CH}$
$\sin CAD = \frac{HE}{AH}$
$\sin ABE = \frac{HF}{BH}$
$\sin BCF = \frac{HD}{CH}$

Thus our previous expression can be rewritten as

$\frac{HF}{AH}\cdot \frac{HD}{BH}\cdot \frac{HE}{CH} = \frac{HE}{AH}\cdot \frac{HF}{BH}\cdot \frac{HD}{CH}.$

This is obviously true so we conclude that the altitudes of a triangle are concurrent.

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Orthocenter

Proof that the altitudes of a triangle are concurrent