Question

In a triangle ABC, DE parallel BC and AD = ( x+3) , DB= (3x +9) , AE=x and CE = (3x+4),then find x​

Answer 1

Answer:

x = -3

Step-by-step explanation:

In triangle ABC,

DE parallel BC

AD = ( x + 3) ,

DB = (3x + 9) ,

AE = x  

CE = (3x + 4)

AD/DB = AE/CE (Thales Theorem)

x + 3/3x + 9 = x/3x + 4 (cross multiply)

(x + 3)*(3x + 4) = x*(3x + 9)

3x^2 + 4x + 9x + 12 = 3x^2 + 9x

3x^2 - 3x^2 + 4x + 9x - 9x + 12 = 0

4x + 12 = 0

4x = -12

x = -12/4

x = -3