Question

in teangle ABC, DE||AB. if AD=2x, DC=x+3,BE=2x-1 and CE=x, then find the value of x

Answer 1

In triangle ABC
DE||AB
According to Thales theorem
DC/ AD = CE/BE
x+3/2x=x/ 2x-1
On cross multiplying we get
x+3(2x-1)=x(2x)
2x^2+x+6x-3=2x^2
2x^2-2x^2-x+6x-3=0
5x-3=0
5x=3
x=3/5

Answer 2

AD = 2x ,DC = x + 3, BE =2x - 1 ,CE = x

Using Thales Theorem

$\bf\huge\frac{CD}{AD} = \frac{CE}{BE}$

$\bf\huge\frac{x+3}{2x} = \frac{x}{2x-1}$

5x = 3

$x=\bf\huge\frac{3}{5}$