Question

P and Q are mid-points of the sides CA and CB respectively of a
∆ABC, right angled at C. Prove that 4AQ2=4AC2+BC2

Answer 1

Answer:

Step-by-step explanation:

In ACB, C = 90o, using Pythagoras theorem,

AB2 = AC2 + BC2 ………………… (i)

In ACQ, C = 90o, using Pythagoras theorem, AQ2 = AC2 + CQ2  

4AQ2 = 4AC2 + BC2 ………………..(ii)

In PCB, C = = 90o, using Pythagoras theorem,

PB2 = PC2 + BC2

PB2 =  

4PB2 = AC2 + 4BC2 ………………(iii)

From (ii) and (iii), 4(AQ2 + PB2) = 5AC2 + 5BC2 = 5(AC2 + BC2) = 5AB2

[Using (i)]

Answer 2

Answer:

Since ∆AQC is a right triangle right angled at C.

AQ^2 = AC^2 +QC^2

4AQ^2 = 4AC^2 +4QC^2

4AQ^2 = 4AC^2 + (2QC)^2

4AQ^2 = 4 AC^2 + BC^2

Hope it is helpful