A circle is touching the side BC of a △ABC at the point P and touching AB and AC produced at points Q and R respectively.
Prove that AQ=21 (Perimeter of △ABC).
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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 △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 △ABC. Therefore, 2AQ is equal to the perimeter of △ABC. Dividing by 2 gives the desired result.