Three equal products at the orthocentre

Idea. The altitudes turn out to be the radical axes of three circles, hence their common point $H$ has equal powers.

The circles. Let $\omega_{AB},\omega_{BC},\omega_{CA}$ be the circles with diameters $AB$, $BC$, $CA$. Since $\angle ADB=\angle ADC=90^\circ$, $D$ lies on $\omega_{AB}$ and on $\omega_{CA}$; similarly $E$ lies on $\omega_{AB}$ and $\omega_{BC}$, and $F$ lies on $\omega_{BC}$ and $\omega_{CA}$.

Radical axes. The circles $\omega_{AB}$ and $\omega_{CA}$ both pass through $A$ and $D$, so their radical axis is the line $AD$. Likewise the radical axis of $\omega_{AB},\omega_{BC}$ is $BE$, and that of $\omega_{BC},\omega_{CA}$ is $CF$. These three lines meet at $H$, which therefore has the same power with respect to the three circles.

Computing the power. The line $AD$ meets $\omega_{AB}$ at $A$ and $D$, so the power of $H$ with respect to $\omega_{AB}$ is $HA\cdot HD$. The line $BE$ meets $\omega_{AB}$ at $B$ and $E$, so it is also $HB\cdot HE$. Likewise $HB\cdot HE=HC\cdot HF$. Therefore
$$HA\cdot HD=HB\cdot HE=HC\cdot HF.\ \blacksquare$$

Remark. The three altitudes concurring is itself a consequence of the radical centre theorem.

Lesson. Circles on diameters give right angles for free; radical axes then produce concurrency and equal products.