Answer :

**Given:** In ΔABC,

AB = AC = BC

**To prove:** 3 AB^{2} = 4 AD^{2}

**Theorem Used:**

1) If corresponding side and two angles of two triangles are equal the triangles are said to have SAS congruency.

2) Pythagoras theorem:

In a right-angled triangle, the squares of the hypotenuse is equal to the sum of the squares of the other two sides.

**Proof:**

Let AD ⊥ BC

In ΔADC and ΔADB

AB = AC (given)

∠B = ∠C (each 60°)

∠ADB = ∠ADC (each 90°)

∴ ΔADC ≅ ΔADB

⇒ BD = DC

As ΔADB is a right triangle right angled at D.

By Pythagoras theorem,

AB^{2} = AD^{2} + BD^{2}

⇒ AB^{2} = AD^{2} + (1/2 BC)^{2}

⇒ AB^{2} = AD^{2} + (1/4 BC^{2})

As BC = AB

⇒ AB^{2} = AD^{2} + (1/4 AB^{2})

⇒ 3 AB^{2} = 4 AD^{2}

Hence proved.

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