Q. 7

# In ΔABC, AB >

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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