Question

in the adjacent figure, a circle touches the side BC , ∆ABC At Point P . all extension, side AB and AC Touch the circle at point Q and R respectively.
Prove that :- AQ =½(perimeter of ∆ABC) .
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Answer 1

Answer:

Step-by-step explanation:

FIGURE IS IN THE ATTACHMENT

Given:

A Circle touches the side BC of ∆ABC at P and extended sides AB  and AC at Q and R.

To Prove:AQ = ½ (BC +CA + AB).

Proof:

Tangents drawn from an external point to the circle are equal.

BP = BQ  ………………....(1)

CP = CR  ……………...….(2)

AQ = AR  ……………...….(3)

Perimeter of ∆ABC = AB + BC + CA

= AB +( BP+ CP) + AC     (BC = BP+CP)

= (AB + BQ) +(CR+AC)    (from eq (2) and (3))

= AQ +AR

= AQ + AQ            (From eq 1)

AB + BC + CA= 2AQ

1/2(AB + BC + CA)= AQ

AQ= 1/2(Perimeter of ∆ABC)

AQ = 1/2(AB + BC + CA)

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