The sum of the lengths of the sides AB + BC + AC is the perimeter of ABC. Therefore, 2AQ is equal to the perimeter of ABC. Dividing by 2 gives the desired result.

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

Answer: $AQ = \frac{1}{2} (\text{Perimeter of } \triangle ABC)$

Step-by-step solution

Step 1: Identify tangent properties

When a circle touches the sides of a triangle (or its extensions), the lengths of tangents drawn from an external point to the circle are equal. Here, from point A, tangents AQ and AR are equal. From point B, tangents BQ and BP are equal. From point C, tangents CR and CP are equal.

Step 2: Express AQ in terms of triangle sides

We can express the length of AQ as the sum of AB and BQ. Similarly, AR can be expressed as the sum of AC and CR. Since AQ = AR, we can use this relationship.

Step 3: Substitute tangent equalities

Substitute the tangent equalities $BQ = BP$ and $CR = CP$ into the expressions for AQ and AR. This allows us to relate AQ and AR directly to the sides of $\triangle ABC$ and the segments on BC.

Step 4: Combine expressions for AQ and AR

Since $AQ = AR$ , we can add the two expressions for AQ and AR. This gives us $2AQ = (AB + BP) + (AC + CP)$ . Rearranging the terms, we get $2AQ = AB + AC + (BP + CP)$ . Since $BP + CP = BC$ , we can substitute this to get $2AQ = AB + AC + BC$ .

Step 5: Relate to perimeter of triangle ABC

The sum of the lengths of the sides $AB + BC + AC$ is the perimeter of $\triangle ABC$ . Therefore, $2AQ$ is equal to the perimeter of $\triangle ABC$ . Dividing by 2 gives the desired result.

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