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= 1/2 perimeter of A ABC.
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Answer 1

Answer:

Since length of tangents from an external point to a

circle are equal,

At A, AX = AY (1)

At B, BX = BZ (2)

At C, CY = CZ (3)

Perimeter of ABC ,

p = AB +AC+ BC

BZ+CZ=BX+CY

BC=AX-AB+AY-AC

AB+BC+AC=AX+AX

Thus AX = 2  PERIMETER

HENCE PROVED...

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