parmenides51 wrote: From the foot of one altitude of the acute triangle, perpendiculars are drawn on the other two sides, that meet the other sides at $P$ and $Q$. Show that the length of $PQ$ does not depend on which of the three altitudes is selected. This length is nothing but $\dfrac{abc}{4R^2}=\dfrac{\Delta}{R}$

It is clear that $APDQ$ is cyclic hence by the sine rule we have, $$\dfrac{PQ}{sin\angle A}=\dfrac{AP}{sin\angle AQP}=\dfrac{\dfrac{AP}{1}}{\dfrac{AP}{AD}}=AD$$So $PQ=ADsin\angle A$ and we have to prove that $ADsin\angle A= BEsin\angle B$ where $E$ is the perpendicular from $B$ to side $AC$ which is true since, $\dfrac{AD*BE}{AB}=\dfrac{AD*BE}{AB}$