Question

D and E are points on the sides AB and AC respectively of a ABC such that DE || BC. Find the value of x,
when (i) AD = x cm, DB = (x - 2) cm, AE = (x + 2) cm and EC = (x - 1) cm.
(ii) AD = 4 cm, DB = (x - 4) cm, AE = 8 cm and EC = (3x - 19) cm.
(iii) AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x + 4) cm and EC = 3x cm.

solve it using thales theorem​

Answer 1

Step-by-step explanation:

Hope this helps and you can tell me anything.

Answer 2

$\frac{AD}{DB} = \frac{ AE }{EC} \\ \frac{x}{(x - 2)} = \frac{(x + 2)}{(x - 1)} \\ ( x- 2)(x + 2) = x(x - 1) \\ {x}^{2} + 2x - 2x - 4 = {x}^{2} - x \\ {x}^{2} - 4 = {x}^{2} - x \\ {x}^{2} - x - {x}^{2} - x \\ x - 4\\ x = 4$