Question

In an equilateral triangle ABC, D is a point on side BC such that 4BD = BC. Prove that 16$A{D}^2$ = 13 $B{C}^2$.

Answer 1

ABC is an equilateral triangle in which $BD=\frac{BC}{4}$

Drawn AE perpendicular to BC

Since AE is perpendicular to BC then $BE=EC=\frac{BC}{2}$ [In equilateral triangle perpendicular drawn from vertex to base bisects the base]

In $△AED$

$A{D}^2=A{E}^2+D{E}^2......\left(1\right)$

In $△AEB$

$A{B}^2=A{E}^2+B{E}^2......\left(2\right)$

Putting the value of $A{E}^2$ from (2) in (1), we get

$A{D}^2=A{B}^2-B{E}^2+D{E}^2$

$A{D}^2=B{C}^2-{\left(\frac{BC}{2}\right)}^2+(BE-BD{)}^2$ [As $△ABC$ is an equilateral triangle]

$A{D}^2=B{C}^2-\frac{B{C}^2}{4}+{(\frac{BC}{2}-\frac{BC}{4})}^2$

$A{D}^2=B{C}^2-\frac{B{C}^2}{4}+\frac{B{C}^2}{16}$

$A{D}^2=\frac{16B{C}^2-4B{C}^2+B{C}^2}{16}$

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

$16A{D}^2=13B{C}^2$