Question

P and q are points on the sides ca and cb respectively of triangle abc right angled at c prove that aq^2 + bp^2 = ab^2+pq^2

Answer 1

in triangle abc measure angle C equals to 90 then AQ square = AC square + CQ square
in triangle PBC measure angle C equals to 90 then PB square = PC square + CB square
and in triangle abc measure angle C equals to 90 and a b square =AC square + BC square and in triangle pqc measure angle C equals to 90 and PQ square =PC square +cq square and then by plusing a b square + PQ square you get the answer

Answer 2

Given - In △ABC ,

angle C = 90°

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)

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{proved}