Answer :

From the given information, we construct the figure as:


As AM is perpendicular BC, ΔACM and ΔABM are right triangle.


By using Pythagoras Theorem, we get,


AC2 = CM2 + AM2 ...... (1)


Also, AB2 = BM2 + AM2 ...... (2)


Eliminating AM2 from equations (1) and (2), we get,


AC2 – CM2 = AB2 – BM2


AC2 – AB2 = CM2 – BM2


By using the identity a2 – b2 = (a + b)(a – b) on above equation,


AC2 – AB2 = (CM + BM)(CM – BM)


AC2 – AB2 = BC(CM – (BC – CM))


AB2 – AC2 = BC(2CM – BC)


As D is mid point of BC,


AB2 – AC2 = BC(2CM – 2BD)


AB2 – AC2 = 2BC(CM – BD)


AB2 – AC2 = 2BC(CM – CD)


AB2 – AC2 = 2BC(–DM)


AB2 — AC2 = 2BC.DM


Hence, proved.


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