Question

A circle is touching the side \[BC\] of \[\vartriangle ABC\] at \[P\] and touching \[AB\] and \[AC\] produced at \[Q\] and \[R\] respectively. Prove that \[AQ = \dfrac{1}{2}\](perimeter of \[\vartriangle ABC\] )

Answer 1

Hint: Here we use the property of tangent to the circle which states that from the same external point, the tangent segments to a circle are equal. Then we write the pair of line segments that are equal using this property and substitute them in the formula for the perimeter of the triangle.
* Tangent to a circle is a line that touches the circle at one point.
* Perimeter of the triangle is the sum of the lengths of all three sides.

Complete step-by-step answer:
We take three points \[A,B,C\] external to the circle and use the tangent property on them.
Looking at the figure, we can say that \[AQ\] and \[AR\] are tangents from point \[A\] to the circle.
Since, the tangent segments to a circle are equal from the same external point.
Therefore, \[AQ = AR\] \[...(i)\]
Similarly, we can say that \[BP\] and \[BQ\] are tangents from point \[B\] to the circle.
Since, the tangent segments to a circle are equal from the same external point.
Therefore, \[BP = BQ\] \[...(ii)\]
And we can say that \[CP\] and \[CR\] are tangents from point \[C\] to the circle.
Since, the tangent segments to a circle are equal from the same external point.
Therefore, \[CP = CR\] \[...(iii)\]
Since, point \[B\] lies on the line segment \[AQ\], so we can write \[AQ = AB + BQ\]
Similarly, point \[C\] lies on the line segment \[AR\], so we can write \[AR = AC + CR\]
Substitute the values of \[AQ = AB + BQ\] and \[AR = AC + CR\] in equation \[(i)\] \[AQ = AR\]
\[
  AQ = AR \\
  AB + BQ = AC + CR \\
 \]
Now substitute the values of \[BP = BQ\]and \[CP = CR\] from equation \[(ii)\] and \[(iii)\]
\[AB + BP = AC + CP\] \[...(iv)\]
Calculating the perimeter of the triangle.
Perimeter of \[\vartriangle ABC = AB + BC + AC\]
Substitute the value of \[BC = BP + PC\] as point \[P\] lies on \[BC\].
Perimeter of \[\vartriangle ABC = AB + (BP + PC) + AC\]
                                      \[ = (AB + BP) + (PC + AC)\]
                                      \[ = (AB + BP) + (AB + BP)\] { since, \[AC + CP = AB + BP\] }
                                      \[ = 2(AB + BP)\]
                                      \[ = 2(AB + BQ)\] { since, \[BP = BQ\] }
                                      \[ = 2AQ\] { since, \[AQ = AB + BQ\] }
Therefore, Perimeter of \[\vartriangle ABC = 2AQ\]
Dividing both sides by \[2\]
\[\dfrac{1}{2}\]Perimeter of \[\vartriangle ABC = \dfrac{{2AQ}}{2}\]
\[\dfrac{1}{2}\]Perimeter of \[\vartriangle ABC = AQ\]

Note:
Students are likely to make mistakes while substituting the values while calculating the perimeter. When we have \[(AB + BP) + (PC + AC)\] then we have to change\[(PC + AC)\] into \[AB + BP\] to make it \[2(AB + BP)\] because we can see in the question LHS has \[AQ\] in it so we can predict to convert which form to which form.