(Q3) BL and CM are medians of a triangle ABC right angled at A.Prove that 4(BL2 + CM2) = 5BC2

ABCML

Solution :

BL and CM are medians of ∆ABC in which ∠A = 90o.

In ∆ABC

BC2 = ( )2 + AC2 (by Pythagorean theorem) ....(1)

In ∆ABL

( )2 = AL2 + AB2

So BL2 = [
AC
2
 ]2 + AB2 (∵ L is the midpoint of AC)
⇒ BL2 =
AC2
4
 + AB2

∴ BL2 = AC2 + 4AB2....(2)

In ∆CMA, CM2 = AC2 + ( )2

⇒ CM2 = AC2 + [
AB
2
 ]2 (∵ M is the midpoint of AB)
⇒ CM2 = AC2 +
AB2
4

CM2 = AC2 + AB2....(3)

On adding (2) and (3), we get

4( 2 + CM2) = 5( 2 + AB2)

∴ 4(BL2 + CM2) = 5BC2   from (1)