P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that AQ2 + BP2 = AB2 + PQ2 [4 MARKS]
Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks
In right- angled triangle ACQ,
AQ2 = AC2 + CQ2....(1)
In right- angled triangle PCB,
PB2 = PC2 + CB2......(2)
In right- angled triangle ABC,
AB2=AC2+BC2.......(3)
In right- angled triangle PQC,
and PQ2=PC2+QC2......(4)
Adding (1) and (2)
⇒AQ2 + BP2 = (AC2 + CQ2) + (PC2 + CB2)
⇒AQ2 + BP2 = (AC2 + BC2) + (PC2 + QC2)
⇒AQ2 + BP2 = AB2 + PQ2
[By (3) and (4)]