Q22 of 56 Page 6

In Δ ABC, ∠ B = 90° and D is the mid-point of BC. Prove that AC2 = AD2 + 3CD2.

Given: In Δ ABC, ∠B = 90° and D is the mid-point of BC.
To Prove: AC2 = AD2 + 3CD2
Proof:
In Δ ABD,
AD2 = AB2 + BD2
AB2 = AD2 - BD2 ……………..(i)
In Δ ABC,
AC2 = AB2 + BC2
AB2 = AC2 - BC2 ……………..(ii)
Equating (i) and (ii)
AD2 - BD2 = AC2 - BC2
AD2 - BD2 = AC2 – (BD + DC)2
AD2 - BD2 = AC2 – BD2 - DC2 – 2BD × DC
AD2 = AC2 - DC2 – 2DC2 (DC = BD)
AD2 = AC2 - 3DC2

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20 In a Δ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4 cm, AE = 8 cm, DB = x – 4
and EC = 3x – 19, find x. 21 In a Δ ABC, AD is the bisector of ∠A, meeting side BC at D. If AC = 4.2 cm, DC = 6 cm, BC = 10 cm,  find AB. 23 ABC is a triangle in which AB = AC and D is a point on the side AC such that BC2 = AC × CD. Prove that BD = BC. 24 The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that
DF × EF = FB × FA.