Question

In traingle ABC, angle A= 90° , P and Q is mid point of AC and AB . prove that: BP square + CQ square = 5PQ square

Answer 1

In right triangle BAP,

BP2 = AB2 + AP2 [Pythagoras theorem] .........(1)

Since Q is the mid point of AB, then AB = 2AQ

So, from (1), we get

BP2 = (2AQ)2 + AP2 = 4AQ2 + AP2 ............(2)



In right triangle QAC,

QC2 = AQ2 + AC2 [Pythagoras theorem] .........(3)

Since P is the mid point of AC, then AC = 2AP

So from (3), we get

QC2 = AQ2 + (2AP)2 = AQ2 + 4AP2 ..............(4)

Adding (2) and (994), we get

BP2 + QC2 = (4AQ2 + AP2) + (AQ2 + 4AP2) = 5AQ2 + 5AP2 = 5(AQ2 + AP2) ..........(5)

In right triangle QAP, we have

AQ2 + AP2 = PQ2 [Pythagoras theorem]

Now from (5), we get

BP2 + QC2 = 5PQ2