Question

∆ABC is an equilateral triangle.Point D is on seg BC such that BD=1/5 ×AC, prove that:25AD²=21 AB²​

Answer 1

Given:

BD =1/5 ×AC

To prove:

25 AD²=21 AB²​

Proof:

1)AB = AC = BC = 10x cm then,

• BD = 1/5 of AC = 2x

• BM = 1/2 of BC = 5x

• DM = BM - BD = 3x

2)In ΔAMB

• AM²+BM = AB² (Pythagoras theorem)

• AM² + (5x)² = (10x)²

• AM²+ 25x² = 100x²

• AM² = 75x²

• AM = 5√3x

3)In ΔAMD

• AM²+DM²=AD²

• (5√3x)²+ (3x)²=AD²

• 75x²+9x²=AD²

• AD=2√21x

4) LHS: 25AD²= 25×84x² = 2100x²

5) RHS: 21AB² = 21 ×100x² = 2100x²

Hence LHS = RHS

Answer 2

Answer:

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