Question

In an equilateral  ΔABC, E is any point on BC such that BE = 1/4 BC. Prove that 16 AE²=13AB²

Answer 1

Join A to mid-point of BC at D. Hence ED = BE = (1/4)BC --(1)
In triangle AED, AE² = AD² + ED² -----------------(2)
In triangle ABD, AD²  = AB² - BD²   --------------(3)
Putting value of AD² from (3) into (2),
AE² = AB² - BD² + ED² = AB² - (BC/2)² + (BC/4)²
as BD = (1/2)BC and ED = (1/4)BC from (1).

OR AE² = AB² - (AB/2)² + (AB/4)² as BC = AB as triangle ABC is equilateral.

Simplifying this , 16AE² = 13AB²

Answer 2

Answer:

Step-by-step explanation:

Join A to mid-point of BC at D. So,ED = BE = (1/4)BC --(1

In triangle AED, AE² = AD² + ED² ------(2

In triangle ABD, AD²  = AB² - BD²   ---(3

Putting value of AD² from (3) into (2),

AE² = AB² - BD² + ED² = AB² - (BC/2)² + (BC/4)²

as BD = (1/2)BC and ED = (1/4)BC from (1)....,,,

So we get ....16AE² = 13AB²