Question

ABC is a triangle.A circle touches the sides AB and AC produced and side BC at X Y abd Z respectively.Show that AX= 1/2 perimeter of triangle ABC

Answer 1

2x+1\2=180
2x+1=180x2
3x=360
x=120

Answer 2

Answer:

Given data:

ABC is a triangle

A circle touches the sides of the triangle AB produced at X, AC produced at Y and BC at Z.

To show: AX = ½ * perimeter of triangle ABC

We know that the lengths of the tangents drawn from an external point to a circle are equal.

Therefore, from the figure below, we get

BX = BZ, AX = AY & CZ = CY ….. (i)

Now,

The perimeter of ∆ ABC  

= AB + BC + AC  

= (AX – BX) + (BZ + ZC) + (AY – CY)

= AX + AY – BX + BZ + ZC – CY

cancelling the similar terms as from (i), we get

= AX + AY

= 2 * AX …… [from (i)]

Hence, AX = ½ * Perimeter of ∆ ABC