Question

A circle touches the side BC of triangle ABC at point P. Now it touches the extended side AB at Q and extended side AC at R. Prove that AQ is ${\displaystyle \frac{AB+BC+AC}{2}}$ .

Answer 1

Hint: Now we know that the length of tangents from the same points are equal. Hence we will get three equations suggesting 3 pairs of triangles are equal. Now Consider the perimeter that is AB + BC + AC. Now we can write BC as BP + PC and AC as AR – CR. Now we will substitute the values from the result in the obtained equation and hence prove the required result.

Complete step by step answer:
Now we are given that the circle touches the line BC at point P. Then it touches the extended side AB at point Q and extended side AC at point R.
Let us draw the figure to understand better.

Now When we say a line touches the circle at a point it means that the line is tangent to the circle at that point.
Hence we get AQ and AR are tangents to the circle at points Q and R. Similarly BP and BQ are tangents to the circle at point P and Q and CP and CR are tangents to the circle at P and R respectively.
Now we know that the length of tangents drawn from the same points are equal.
Hence using this we get
$\begin{array}{rl}& AQ=BC.............\left(1\right)\\ & BP=BQ............\left(2\right)\\ & CP=CR..............\left(3\right)\end{array}$
Now Consider AB + BC + AC.
From the figure we can say that BC = BP + PC and AC = AR – AQ.
Hence we get,
AB + BC + AC = AB + BP + PC + AQ – CR.
Now from equation (1), equation (2) and equation (3) we get,
AB + BC + AC = AB + BQ + CR + AQ – CR
Now from the figure we can say that AB + BQ = AQ
Hence we get $AB+BC+AC=2AQ$
Hence we get $AQ={\displaystyle \frac{AB+BC+AC}{2}}$ .
Hence, proved.

Note: Now we know that when we say the tangents are of equal length they must be tangents from the same point. Also when substituting the values from the result substitute smartle so that we get the required result. Hence as we needed AQ we tried to write everything in the form of AQ.