In an equilateral triangle ABC,D is a point on side BC such that 4BD = BC prove that 16AD2 - Maths - Triangles

Question

In an equilateral triangle ABC,D is a point on side BC such that
4BD = BC prove that 16AD2 = 13BC2

Varun.Rawat · Accepted Answer

ABC is an equilateral triangle in which BD = BC / 4. Draw AE perpendicular to BC.

Since AE is perpendicular to BC, then BE = EC = BC / 2 [ In equilateral Δ, ⊥ drawn from vertex to base bisects the base]

In Δ AED, AD²= AE²+ DE²[Pythagoras Theorem] ..............(1)

In Δ AEB, AB²= AE²+ BE²[Pythagoras Theorem] ............(2)

Putting the value of AE²from (2) in (1), we get

AD²= AB²- BE² + DE²

AD²= BC²- (BC / 2)²+ (BE - BD)²[ As, ABC is an equilateral Δ, AB = BC = CA ]

$= {B C}^{2} - \frac{{B C}^{2}}{4} + {[\frac{B C}{2} - \frac{B C}{4}]}^{2} = {B C}^{2} - \frac{{B C}^{2}}{4} + \frac{{B C}^{2}}{16} = \frac{16 {B C}^{2} - 4 {B C}^{2} + {B C}^{2}}{16} = \frac{13 {B C}^{2}}{16}$

$S o, {A D}^{2} = \frac{13 {B C}^{2}}{16}$

⇒ 16 AD²= 13 BC²

In an equilateral triangle ABC,D is a point on side BC such that 4BD = BC. prove that 16AD2 = 13BC2

In an equilateral triangle ABC,D is a point on side BC such that 4BD = BC. prove that 16AD² = 13BC²