Q. 64.7( 10 Votes )

# ABC is an equilat

Answer :

Given: AB = BC = CA = x

∠ADC = ∠ADB = 90°

To prove: AB^{2} + BC^{2} + CA^{2} = 4AD^{2}

Proof:

We know that in an equilateral triangle perpendicular from any vertex on opposite side bisects it.

So, BD = DC = 1/2BC ……………… (1)

By applying Pythagoras theorem in ΔABD, we get,

AB^{2} = AD^{2} + BD^{2} [∵ H^{2} = P^{2} + B^{2}]

By substituting BD from eqn. (1) we get,

AB^{2} = AD^{2} + (1/2BC)^{2}

⇒ AB^{2} = AD^{2} + 1/4BC^{2}

⇒ 4AB^{2} = 4AD^{2} + BC^{2}

⇒ 4AB^{2} - BC^{2} = 4AD^{2}

⇒ 4AB^{2} - AB^{2} = 4AD^{2} [Given: AB = BC]

⇒ 3AB^{2} = 4AD^{2}

⇒ AB^{2} + BC^{2} + CA^{2} = 4AD^{2} [∵ AB = BC = CA]

Hence proved.

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