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                 

AC = BC               [Given]

CD = CD              [Common]

AD = DB              [D is the mid point]

⇒ ΔADC ≅ ΔADB             [By SSS Congruency Criterion]

⇒ ∠ADC = ∠ADB           [Corresponding parts of congruent triangles are equal]

Also, ∠ADC + ∠ADB = 180°      [Linear Pair]

Hence, ADC and ADB are right triangles, ∴ By Pythagoras theorem

AC2 = CD2 + AD2            

BC2 = CD2 + BD2            

Adding  and , we have

AC2 + BC2 = AD2 + BD2 + 2CD2

⇒ AB2 = AD2 + BD2 + 2CD2

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

⇒ 2AD2 = 2CD2

⇒ AD2 + BD2 = 2CD2    [∵ AD = BD]

Hence, Proved!

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