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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).

 

 

- Daisy

 Aug 9, 2018
edited by dierdurst  Aug 22, 2018
 #1
avatar+994 
+5

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

 Aug 9, 2018
 #2
avatar+395 
+5

Thank you very much, this solution is very clear!

 

- Daisy

dierdurst  Aug 9, 2018
edited by dierdurst  Aug 22, 2018

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