Question

in the figure of triangle ABC DE||AB, AD= 2x ,DC = X + 3 , BE= 2 x -1 and CE = X, find the value of x

Answer 1

this question is solved with the help of Thales theorem or Basic proportionality theorem

Answer 2

By B.P.T theorem.......

$\frac{ad}{dc} = \frac{be}{ec} \\ \: \:$

$\frac{2x}{x + 3} = \frac{2x - 1}{x} \\ \: \\ \\ 2x(x) = (2x - 1)(x + 3) \\ \\ 2x {}^{2} = 2x {}^{2} + 6x - x - 3 \\ \\ 6x - x - 3 = 0 \\ \\ 5x - 3 = 0 \\ \\ 5x = 3 \\ \\ x = \frac{3}{5}$