Question

Seg Ad perpendicular to bc and prove AB^2 + AC^2 =2AM^2 + 2BM^2
(take ^2 as square)

Answer 1

Step-by-step explanation:

Pythagoras in obtuse angle M

AB^2= AM^2 + BM^2 +2BM×MD.--------(1)

Pythagoras in acute angle M

AC^2= AM^2 + MC^2-2MC×MD.-–------(2)

Adding 1 and 2

AB^2 + AC^2 = 2AM^2 + 2BM^2 {Because BM = MC as AM is median.

This above result us known as apollonius thereom.

Mark brainliest!!