Question

P and Q are points on the sides CA and CB of a triangle ABC, right angled at C. Prove that
(AQ²+BP²) =(AB²+PQ²)​

Answer 1

$\huge \mathfrak{ \underline{answer - }}$

Given - A $\triangle \: ABC$In which $\angle \: C = 90 \degree$P and Q are points on CA and CB respectively.

To prove - (AQ² +BP ²)= (AB ²+PQ ²).

Proof - From right $\triangle \: ACQ$

AQ²= (AC² +CQ ²). .......(1). [by Pythagoras' theorem]

From right $\triangle \: BCP$

BP²= (BC ²+CP²). ........( 2). [by Pythagoras theorem]

From right $\triangle \: ACB$

AB² = (AC²+BC². .......... (3). [by Pythagoras theorem]

From right $\triangle \: PCQ$

PQ²= (CQ²+CP²). .......... (4). [by Pythagoras theorem]

From (1)&(2),we get

(AQ²+BP²) = (AC²+BC²) +(CQ²+CP²)

= (AB²+PQ²). [Using (3)&(4)]

Hence, (AQ²+BP²) = (AB²+PQ²)

$\huge \underline{ \underline \bold {proved}}$