In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC =

1.	In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.
Answer» In ∆ABC, seg BD bisects ∠ABC. [Given] ∴ AB/BC = AD/CD [Property of angle bisector of a triangle] ∴ x/(x + 5) = (x - 2)/(x + 2) ∴ x(x + 2) = (x – 2)(x + 5) ∴ x² + 2x = x²+ 5x – 2x – 10 ∴ 2x = 3x – 10 ∴ 10 = 3x – 2x ∴ x = 10

In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Answer»

In ∆ABC, seg BD bisects ∠ABC. [Given]

∴ AB/BC = AD/CD [Property of angle bisector of a triangle]

∴ x/(x + 5) = (x - 2)/(x + 2)

∴ x(x + 2) = (x – 2)(x + 5)

∴ x² + 2x = x²+ 5x – 2x – 10

∴ 2x = 3x – 10

∴ 10 = 3x – 2x

∴ x = 10

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In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

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