Q. 11

# ABC is and isosce

Given: ABC is an isosceles triangle. So, AC = BC

C = right angle

To prove: AD2 + DB2 = 2CD2

Proof:
By Pythagoras Theroem,

In ΔABC, We have

AB2 = AC2 + BC2                 [1]

AC = BC               [Given]

CD = CD              [Common]

AD = DB              [D is the mid point]

AC2 = CD2 + AD2            [2]

BC2 = CD2 + BD2            [3]

Adding [2] and [3], we have

AC2 + BC2 = AD2 + BD2 + 2CD2

⇒ AB2 = AD2 + BD2 + 2CD2

[∵ D is mid-point of AB, AD = BD and AB = 2AD = 2BD]

Hence, Proved!

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