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]

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!