Question

A circle touches the sides BC of a $\Delta ABC$ at P and touches AB and AC when produced at Q and R respectively as shown in the figure. Show that $AQ=\frac{1}{2}$ (Perimeter of $\Delta ABC$)
If true answer ${}^{\prime }{0}^{\prime }$ else ${}^{\prime }{1}^{\prime }$

Answer 1

Given: A circle is touching side BC of $\Delta ABC$ at P and touching AB and AC when produced at Q and R
respectively.

To prove: $AQ=\frac{1}{2}$ (perimeter of $\Delta ABC$)

Proof: $AQ=AR$ ...(i)
$BQ=BP$ .....(ii)
$CP=CR$ ......(iii)
[Tangent drawn from an external point to a circle are equal]

Now, perimeter of $\Delta ABC=AB+BC+CA$

$=AB+BP+PC+CA$

$=(AB+BQ)+(CR+CA)$

[From (ii) and (iii)]
$=AQ+AR=AQ+AQ$ [From (i)]

$AQ=\frac{1}{2}$ (perimeter of $\Delta ABC$).