In ΔABC, AB > AC, D is the mid–point of . such that B—M—C. Prove that AB² — AC² = 2BC . DM.

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.