Question

triangle ABC is a right angled triangle angle A = 90, prove that BL ^2 + CM^2 = 5/4BC^2 where BC and CM are median​

Answer 1

Given,

ABC is a triangle.

∠A = 90°

BL and CM are median

To prove,

BL² + CM² = 5/4BC²

Solution,

In ∆ BAL,

$\sf BL^2 = AL^2 + AB^2$

According to the question,

BL and CM are median so,

AL = LC = 1/2AC

AM = BM = 1/2AB

$\sf BL^2 = \bigg( \dfrac{1}{2} AC \bigg)^2 + AB^2$

$\sf BL^2 = \dfrac{{AC}^{2} }{4} + AB^2$

$\sf BL^2 = \dfrac{{AC}^{2} + 4AB^2 }{4} \: \: \: ...(1)$

Now,

In ∆ AMC

$\sf CM^2 = AC^2 + AM^2$

$\sf CM^2 = AC^2 +\bigg( \dfrac{1}{2} AB \bigg)^2$

$\sf CM^2 = AC^2 + \dfrac{ {AB}^{2} }{4}$

$\sf CM^2 = \dfrac{4AC^2 + {AB}^{2} }{4} \: \: \: ...(2)$

According to the question,

$\sf LHS = BL^2 + CM^2$

$\sf = \dfrac{{AC}^{2} + 4AB^2 }{4} + \dfrac{4AC^2 + {AB}^{2} }{4}$

$\sf = \dfrac{{AC}^{2} + 4AB^2 +4AC^2 + {AB}^{2} }{4}$

$\sf = \dfrac{5AB^2 +5AC^2 }{4}$

$\sf = \dfrac{5(AB^2 +AC^2) }{4}$

$\sf = \dfrac{5}{4} BC^2 = RHS$

LHS = RHS

$\sf BL^2 + CM^2 = \dfrac{5}{4} BC^2$

Hence proved !!