Question

2) ∆ABC is an equilateral triangle
Point D is on seg BC such that BD=1/5 BC. Prove
That 25 AD²=21AB²​

Answer 1

Answer:

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