The point of intersection of the common chords of three circles described on the three sides of a triangle as diameter is

Let $ABC$ be a triangle and let $AD, BE$ and $CF$ be its altitudes. Then,

$\mathrm{\angle }ADB=\pi /2\text{ and }\mathrm{\angle }AEB=\pi /2$ Therefore, the circle on $AB$ and $AC$ as diameters passes through $D$. Consequently,$AD$ is the common chord of the circles on $AB$ and $AC$ as diameters.

Similarly, the other two altitudes $BE$and $CF$ are the other two common chords. Thus, the point of intersection of the common chords is the point of intersection of $AD, BE$ and$CF$i.e. the orthocentre of $∆$ $ABC.$