Question

in figure given below De parallel to BC find the value of x..pls guys give answer...no spam right answer I want ​

Answer 1

Given : A  Triangle ABC in which  DE ║ BC   AD = x  , BD = x + 1 , AC = x + 3, CE = x + 5

To find :  Value of x

Solution:

DE ║ BC

=> AD/BD  = AE/CE

=> x/(x +1 )  = (x + 3)/(x + 5)

=> x(x + 5)  = (x + 3)(x + 1)

=> x² + 5x  = x² + 4x + 3

=> x = 3

Value of x = 3

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Answer 2

Answer:

DE II BC

Angle E =Angle B= 50° (Transversal)

Angle ADE =180°-95°=85°

Therefore, Angle x+50°+85°=180°

Angle x=180°-135°=45°

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