Question

D is a point on side BC of an equilateral triangle ABC such that DC = ¼ BC Prove that AD2 = 13CD2 OR ABC is a triangle right angled at B, in which D & E trisects BC prove that 8 AE2 = 3AC2 + 5 AD2​

Answer 1

$\huge \mathcal \red {\underline{\underline{A}}} \huge \mathcal \green {\underline{\underline{N}}} \huge \mathcal \pink {\underline{\underline{S}}} \huge \mathcal \blue {\underline{\underline{W}}} \huge \mathcal \orange {\underline{\underline {E}}} \huge \mathcal \purple {\underline{\underline{R}}} ✧$

Given :-

ABC is a triangle right angled at B, in which D & E trisects BC.

Therefore BD = DE = EC = x (say)

To prove :-

8AE² = 3AC² + 5AD²

Proof :-

In right triangle ️ ABC

AC² = ACB² + BC²

AC² = AB² + (3x)²

AC² = AB² + 9x²........(1)

In right triangle ️ ABD

AD² = AB² + BD²

AD² = AB² + x²........(2)

In right triangle ️ ABE

AE² = AB² + BE²

AE² = AB² + (2x)²

AE² = AB² + 4x²........(3)

Consider RHS

3AC² + 5AD²

=3(AB² + 9x²) + 5(AB² + x²) [using (1) and (2)]

= 8AB² + 32x²

= 8(AB² + 4x²)

= 8AE² [using (3)]

Hence proved.

$\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}$