Question

answer this question

Answer 1

Answer:

Step-by-step explanation:

IN ΔABM

BY PYTHAGORAS THEOREM

AB²=BM²+AM²..........(1)

IN ΔAMD

AD² = AM²+MD².............(3)

IN ΔAMC

AC²=AM²+MC²..............(2)

ADD (1) AND (2)

AB² + AC² = BM²+AM²+AM²+MC²

                 = 2AM²+(BD-MD)²+(CD+MD)²

                 = 2AM²+ BD² +MD² - 2 x BD x MD + CD² +MD²+ 2 x CD x MD

                 = 2AM²+2MD² + BD² - 2 x CD x MD +CD² + 2 xCD xMD

                = 2 AD² + 1/4 x BC² + 1/4 x BC²

                = 2AD² + 2 x 1/4 x BC²

               = 2AD² + 1/2 BC²

∴ AB²+AC² = 2AD² + 1/2 BC²

HENCE PROVED

Answer 2

Hope it helps you ......