Question

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${AQ}^2$ + ${BP}^2$ = ${AB}^2$ + ${PQ}^2$ [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks




In right- angled triangle ACQ,

${AQ}^2$ = ${AC}^2$ + ${CQ}^2....\left(1\right)$

In right- angled triangle PCB,

${PB}^2$ = ${PC}^2$ + ${CB}^2......\left(2\right)$

In right- angled triangle ABC,

${AB}^2={AC}^2+{BC}^2.......\left(3\right)$

In right- angled triangle PQC,

and ${PQ}^2={PC}^2+{QC}^2......\left(4\right)$

Adding (1) and (2)

$\Rightarrow {AQ}^2$ + ${BP}^2$ = (${AC}^2$ + ${CQ}^2$) + (${PC}^2$ + ${CB}^2$)

$\Rightarrow {AQ}^2$ + ${BP}^2$ = (${AC}^2$ + ${BC}^2$) + (${PC}^2$ + ${QC}^2$)

$\Rightarrow {AQ}^2$ + ${BP}^2$ = ${AB}^2$ + ${PQ}^2$

[By (3) and (4)]