Q. 20

# In ΔABC, ∠B = 90° and D is the midpoint of BC. Prove that AC2 = AD2 + 3CD2. Given: B = 90° and D is the midpoint of BC .i.e. BD = DC

To Prove: AC2 = AD2 + 3CD2

In ∆ABC, using Pythagoras theorem we have,

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AB)2 + (BC)2 = (AC)2

(AB)2 + (2CD)2 =(AC)2

(AB)2 + 4(CD)2 =(AC)2

(AD2 – BD2) + 4(CD2) = AC2

[ In right triangle ∆ABD, AD2 =AB2 + BD2 ]

AD2 – BD2 + 4CD2 = AC2

AD2 – CD2 + 4CD2 = AC2

[ D is the midpoint of BC, BD = DC]

or AC2 = AD2 + 3CD2

Hence Proved

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