Question

In the given figure, DE || BC. Find the value of x.​

Answer 1

Answer :-

The value of x is 3.

Explanation :-

From Thales' Theorem,

$\frac{</strong><strong>AD</strong><strong>}{</strong><strong>DB</strong><strong>} = \frac{</strong><strong>AE</strong><strong>}{</strong><strong>EC</strong><strong>} \\ = > \frac{x + 1}{x - 1} = \frac{x + 3}{x} \\ \\ </strong><strong>B</strong><strong>y \: cross \: multiplication</strong><strong>,</strong><strong> \\ \\ x(x + 1) = (x - 1)(x - 3) \\ = > {x}^{2} + x = {x}^{2} + 3x - x - 3 \\ = > {x}^{2} - {x}^{2} + x - 2x = - 3 \\ = > - x = - 3 \\ = > x = 3$

•°• The value of x is 3.

Additionally :-

AD = x + 1 = 3 + 1 = 4

DB = x - 1 = 3 - 1 = 2

AE = x + 3 = 3 + 3 = 6

EC = x = 3

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