In $ \triangle A B C, \angle A $ is obtuse, $ P B \perp A C, $ and $ Q C \perp A B $. Prove that $ A B \times A Q=A C \times A P $.

Given:

In \( \triangle A B C, \angle A \) is obtuse, \( P B \perp A C, \) and \( Q C \perp A B \). 

To do:

We have to prove that \( A B \times A Q=A C \times A P \).
Solution:
 

In $\triangle APB$ and $\triangle AQC$,

$\angle APB=\angle AQC=90^o$

$\angle BAP=\angle CAQ$   (Vertically opposite angles)

Therefore,

$\triangle APB \sim\ \triangle AQC$    (By AA similarity)

This implies,

$\frac{AP}{AQ}=\frac{AB}{AC}$   (Corresponding parts of similar triangles are proportional)

$\Rightarrow AP\times AC=AB\times AQ$

Hence proved.

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Updated on: 10-Oct-2022

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