Question

A circle touches side BC of Triangle ABC from outside
of triangle. Further, extended lines AC and AB
are tangents to the circle at N and M respectively.
Prove that AM =
$\frac{1}{2} \: (perimeter \: of \: triangle \: abc)$
​

Answer 1

Answer:

AB +BC+AC=Perimeter of triangle

AB +BP+PC +AC =P. OF ∆

(AB+BM) +(CN+AC) = P.OF∆. [BM=BP AND CN=CP]

AM+AN=P.OF∆

2AM=P.OF∆. [AM=AN BECAUSE they are TANGENTS]

AM=1/2 P.OF ∆

Step-by-step explanation:

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