Question

In a triangle ABC , B=90 degree and M is the Mid point of BC.Prove that AC=AM²+3BM².

Answer 1

IF IT IS AC^2 THEN THE ANSWER IS

FROM TRIANGLE ABC
              BC = 2BM

FROM TRIANGLE ABC

BY PYTHAGORAS THEOREM 

AC^2 = AB^2+BC^2
I.E AB^2 = BC^2-AC^2     ----(1)

FROM TRIANGLE ABM

AM^2 = AB^2+BM^2       (BY PYTHAGORAS THEOREM)
I.E AB^2 = BM^2-AM^2    ------(2)

FROM (1) & (2)

BC^2-AC^2 = BM^2-AM^2
BC^2-BM^2+AM^2 = AC^2
(2BM)^2-BM^2+AM^2 = AC^2        [BC = 2BM]
4BM^2-BM^2+AM^2 = AC^2
3BM^2+AM^2 = AC^2

AC^2 = AM^2+3BM^2

HENCE PROVED

Answer 2

Answer:

Step-by-step explanation:

given,

BM = MC  (AM is median)

To prove,

AC^2 = AM^2 + 3BM^2

     

Proof,

AC^2 = AB^2 + BC^2

        = AB^2 + ( BM + MC )^2

        = (AM^2 - BM^2)  +  BM^2 + MC^2 + 2BM.MC           (expand)

       

Conveting all MC into BM   (BM=MC) ,

         = AM^2 - BM^2 + BM^2 + BM^2 + 2BM^2

         = AM^2 - BM^2 + 4BM^2

         = AM^2 + 3BM^2

HENCE PROVED