Question

2. A circle is touching the side BC of AABC at X and touching AB and AC produced at P
and Q respectively. Prove that AP=AQ=1/2(Perimeter of AABC). Given AP=10cm
Find the perimeter of AABC.​

Answer 1

Answer:

ANSWER

In the figure shown,

Let the sides,

AB=c

AC=b

BC=a

PC=x

PB=a−x

Now, AQ and AR are two pair of tangents drawn from the same extrernal point A. Also, PC and CR are tangents from point C and PB and QB are tangents from point B.

Using the property that lenght of tangents drawn to a circle from a given external point is always the same, we can say that

PC=CR=x

BP=QB=a−x

AQ=AR

=>AB+BQ=AC+CR

=> c+a−x=b+x

=> x=

2

a−b+c

Now,

AQ=AR=b+x

=> b+

2

a−b+c

=>

2

a+b+c

=> Semiperimeter of Triangle ABC.