Question

In triangle ABC the
points Dand E are on the sides such the sides CA ,CB,respectivelysuch that DE is parallel toAB. AD=2x ,DC=x+3,BE=2x-1 and CE=x then find x

Answer 1

Given AD = 2x
CD = x +3
EC. = x
EB. = 2x - 1
ED || AB
< CAB = < CDE [ corresponding angles]
< C is common
Triangle ABC ~ Triangle CDE
$\frac{x + 3?}{2x?} = \frac{x}{2x - 1?} \\ 2 {x}^{2} - x + 6x - 3 = 2 {x}^{2} \\ x = \frac{3}{5}$