A circle is touching the side BC of $\triangle ABC$ at P and touching the sides AB and AC produced at Q and R respectively. Prove that $AQ = \frac{1}{2} (AB + BC + AC)$ .

Answer: $AQ = \frac{1}{2} (AB + BC + AC)$

Step-by-step solution

Step 1: Identify Tangent Properties

According to the property of tangents drawn from an external point to a circle, the lengths of the tangents are equal. From point A, tangents AQ and AR are drawn to the circle, so $AQ = AR$ . From point B, tangents BQ and BP are drawn, so $BQ = BP$ . Similarly, from point C, tangents CR and CP are drawn, so $CR = CP$ .

Step 2: Express AQ in terms of AB and BQ

From the diagram, we can see that the length of AQ is the sum of the lengths of AB and BQ. This is because Q lies on the line AB produced.

Step 3: Substitute BQ and CR

Using the tangent properties from step 1, we substitute $BQ$ with $BP$ in the expression for $AQ$ . Similarly, we can express $AR$ as $AC + CR$ , and then substitute $CR$ with $CP$ .

Step 4: Add the expressions for AQ and AR

We add the two expressions for $AQ$ and $AR$ obtained in the previous step. This allows us to combine the terms related to the sides of $\triangle ABC$ .

Step 5: Simplify using tangent properties

Since $AQ = AR$ , we can replace $AQ + AR$ with $2AQ$ . We also notice that $BP + CP$ is equal to the length of the side $BC$ . Substituting these into the equation simplifies it.

Step 6: Solve for AQ

Finally, we divide both sides of the equation by 2 to get the desired result, proving that $AQ$ is half the perimeter of $\triangle ABC$ .

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A circle is touching the side BC of △ABC\triangle ABC△ABC at P and touching the sides AB and AC produced at Q and R respectively. Prove that AQ=12(AB+BC+AC)AQ = \frac{1}{2} (AB + BC + AC)AQ=21​(AB+BC+AC).