We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+4
424
2
avatar+391 

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+987 
+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+391 
+5

Thank you very much, this solution is very clear!

 

- Daisy

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

9 Online Users

avatar
avatar