Question

in a triangle ABC.. angle B = 90 degree and M is the mid point of BC.. prov tht AC^2 = AM^2 + 3BM^2

Answer 1

In ∆ABM, AB² = AM²-BM² ...(i)
Now M is the mid-point of BC.
=> BC = 2BM ...(ii)
Now, in ∆ABC, AB² = AC²-BC²
=>AB² = AC²-(2BM)² [From (ii)]
=>AB² = AC²-4BM² ...(iii)

From (i) and (iii), we have
AM²-BM² = AC²-4BM²
AM²-3BM² = AC²

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