A circle is touching the side BC of △ABC at P and touching the sides AB and AC produced at Q and R respectively. Prove that AQ=21(AB+BC+AC).
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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 △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 △ABC.