A geometry problem by Ajay Sambhriya

If AL = $x$ ,

$CL$ $=$ $ED$ $=$ $\sqrt3x$

$\angle$ $EBD$ $=$ $120^\circ$

Apply $Law$ $of$ $Cosines$ on $\triangle EBD$

$x^2 - 2x - 8 = 0$

$\implies$ $x = 4, -2$

As a side length( $x$ ) cannot be negative, $x = 4$ $\implies$ $AB = 2x = 2 \cdot 4 = \color{#3D99F6} 8$

Let each side of Triangle ABC be 'a' CL = ED = √3a/2 , CE = EB = a/2 Angle EBD = 120° Applying sine rule , ED / sin120 = EB / sin EDB Solving for Angle EDB , we get Angle EDB = 30° = Angle BED Therefore , EB = BD
a/2=4 Therefore , a=8

Let CE = EB = x

Since it is equilateral, BL = x and CL = x*sqrt(3)

So ED = x*sqrt(3)

By Law of Cosines (x sqrt(3))^2 = x^2 +4^2 - 2 4 x cos120

3x^2 = x^2 +16 +4*x

2x^2 - 4*x - 16 = 0

x^2 - 2*x - 8 = 0

(x-4)*(x+2) = 0

x = 4 or -2 discard negative value

x = 4 so CB = 2*x = 2 times 4 = 8