Altitudes AP and BQ of an acute triangle ABC intersect at point H. If HP=5 while HQ=2, then calculate (BP)(PC)-(AQ)(QC).

dierdurst
Aug 9, 2018

#1**+3 **

This similarity relationship is derived from the fact that each triangle has one right angle and acute angle in common:

\(\because \angle BQC= \angle BPH= \angle HQA = \angle CPA\\ \because \angle QBC = \angle QBC\ \&\ \angle PAC = \angle PAC \ \& \ \angle BHP=\angle AHQ\)

\(\therefore \triangle BQC\sim \triangle BPH\sim \triangle HQA\sim \triangle CPA\)

From the similarity of corresponding triangles:

\(\frac{HP}{BP}=\frac{HQ}{QA}=\frac{PC}{AP}=\frac{QC}{BQ}\) results in: \(BP\cdot PC-AQ\cdot QC =HP\cdot AP-HQ\cdot BQ\\ =HP(HP+HA)-HQ(HQ+HB)=HP^2-HQ^2=\boxed{21}.\)

GYanggg
Aug 9, 2018