Altitudes $\overline{AP}$ and $\overline{BQ}$ of an acute triangle $\triangle ABC$ intersect at point $H$. If $HP=5$ while $HQ=2$, then calculate $(BP)(PC)-(AQ)(QC)$.
When I first saw this problem, I immediately thought of similar triangles. However, I currently cannot find the solution with them.
Help is greatly appreciated.
On
Hint: since $ABPQ$ is cyclic, it follows from the power of a point that:
$$CP \cdot CB = CQ \cdot CA \;\;\iff\;\; CP \cdot (CP+PB) = CQ \cdot (CQ+QA)$$
Using that triangles $\triangle HPC, \triangle HCQ$ are right, it then follows that:
$$\require{cancel} CP \cdot PB - CQ \cdot QA = CQ^2 - CP^2= (\cancel{CH^2}-HQ^2)-(\cancel{CH^2}-HP^2) = \ldots $$
From the similarity of corresponding triangles: $$\frac{HP}{BP}=\frac{HQ}{QA}=\frac{PC}{AP}=\frac{QC}{BQ}$$ follows: $$BP\cdot PC-AQ\cdot QC =HP\cdot AP-HQ\cdot BQ\\ =HP(HP+HA)-HQ(HQ+HB)=HP^2-HQ^2=21,$$ where the identity $$PH\cdot HA=QH\cdot HB\tag{*}$$ was used.
The last identity can be aquired from the problem solved elsewhere, where it was shown (in notation of the reference) that $HG=2HF$ and $HK= 2 HD$. The identity (*) then follows from the well-known equality for the intersecting chords: $$ AH\cdot HK=CH\cdot HG \Leftrightarrow AH\cdot 2 HD=CH\cdot 2 HF\Leftrightarrow AH\cdot HD=CH\cdot HF. $$