Q. 11

ABC is and isosceles triangle whose C is right angle. If D is mid point of AB, then let us prove that, AD2 + DB2 = 2CD2

Answer :

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]

In ΔADC and ΔADB, we have

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]

⇒ ∠ADC + ∠ADC = 180°

⇒ ∠ADC = ∠ADB = 90°

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

AC2 = CD2 + AD2            [2]

BC2 = CD2 + BD2            [3]

 Adding [2] and [3], we have

AC2 + BC2 = AD2 + BD2 + 2CD2

⇒ AB2 = AD2 + BD2 + 2CD2

⇒ (2AD)2 = AD2 + AD2 + 2CD2              

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

⇒ 4AD2 = 2AD2 + 2CD2

⇒ 2AD2 = 2CD2

⇒ AD2 + BD2 = 2CD2    [∵ AD = BD]

Hence, Proved!

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