Answer :

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

∠C = right angle

To prove: AD^{2} + DB^{2} = 2CD^{2}

Proof:

By Pythagoras Theroem,

In ΔABC, We have

AB^{2} = AC^{2} + BC^{2}_{ } [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

AC^{2} = CD^{2} + AD^{2} [2]

BC^{2} = CD^{2} + BD^{2} [3]

Adding [2] and [3], we have

AC^{2} + BC^{2} = AD^{2} + BD^{2} + 2CD^{2}

⇒ AB^{2} = AD^{2} + BD^{2} + 2CD^{2}

⇒ (2AD)^{2} = AD^{2} + AD^{2} + 2CD^{2}

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

⇒ 4AD^{2} = 2AD^{2} + 2CD^{2}

⇒ 2AD^{2} = 2CD^{2}

⇒ AD^{2} + BD^{2} = 2CD^{2} [∵ AD = BD]

Hence, Proved!

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