Question

ABC is a triangle. A circle touches sides AB and AC produced and side BC at point X, Y and Z respectively. Show that half perimeter of triangle ABC is equal to side AX.

Answer 1

Answer:

We know that the perimeter of the triangle is the sum of the three sides of the triangle.

Now, for ΔABC we have,

Perimeter of ΔABC = AB + AC + BC ….. (1)

From the figure AB can be written as:

AB = AX – BX

Similarly, we can write AC and BC as:

AC = AY – CY

BC = BZ + ZC

Substituting all these values in equation (1) we get:

Perimeter of ΔABC = (AX – BX) + (AY – CY) + (BZ + ZC)

⇒ Perimeter of ΔABC = AX – BX + AY – CY + BZ + ZC …… (2)

We also know that the tangents drawn from an external point are equal in length.

Now, from point B we have the tangents BX and BZ. Hence, we can write:

BX = BZ

Similarly from the point C we have the tangents CZ and CY. So, we can write:

CY = CZ

Now, by substituting all these values in equation (2), we obtain:

Perimeter of ΔABC = AX – BZ + AY – CZ + BZ + ZC

Next, by cancellation of BZ and CZ we obtain:

Perimeter of ΔABC = AX + AY ……. (3)

From the figure we can say that AX and AY are tangents drawn from an external point A and are equal in length. Therefore, we get:

AX = AY

So, now equation (3) becomes:

Perimeter of ΔABC = AX + AX

⇒ Perimeter of ΔABC = 2AX

Next, by cross multiplication,

⇒12 Perimeter of ΔABC = AX

That is, AX = 12 Perimeter of ΔABC

Hence, the proof.