Question

ABC is a triangle. A circle touches sides AB and AC produced and side BC at BC at X , X Y, and Z respectively. Show that

AX 2 = 1 perimeter of TABC .​

Answer 1

Proof:

1) to prove that the perimeter of the triangle ABC is equal to twice of x

2) we have to use the theorem of the tangent of circle which states that the tangents of a circle from a point it will be equal to each other.

3) We can clearly see that  AX, AY ; BZ , BX ; CZ , CY are the pair of the tangent of the circle

so,

AX = AY

BX = BZ

CZ = CY

• Perimeter of ΔABC = AB + BC + AC

                                        = AB + BZ + ZC + AC

                                        = AB + BX + CY + AC

                                        = AX + AY

                                        = AX + AX

                                        = 2 AX

• Hence, Perimeter of ΔABC = 2 AX

   AX = 1/2 Perimeter of ΔABC