Question

A circle touches the side BC of a ∆ ABC at P and the extended sides AB and AC at Q and R, respectively. Prove that AQ = ½ (BC +CA + AB).
(Class 10 Maths Sample Question Paper)

Answer 1

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)

HOPE THIS WILL HELP YOU...

Answer 2

Step-by-step explanation:

The tangent drawn from an external point to the circle are equal

BP=BQ................(1)

CQ=CR................(2)

AP=AR................(3)

perimeter of (ABC) = AB+BC+CA

= AB+BP+CP+CA

= (ab+bq) + ( cr+ca)

= aq+ar

= aq+ aq

= 2aq

1/2 perimeter of abc= aq