CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A circle touches the side $$BC$$ of $$\triangle\ ABC$$ at $$P$$ and touches $$AB$$ and $$AC$$ produced at $$Q$$ and $$R$$ respectively. Prove that $$AQ=\dfrac{1}{2}$$ (perimeter of $$\triangle\ ABC$$).
1330880_1f2322d1048645cb8693f15d7a9dab29.PNG


Solution

Given:A circle touching the side $$BC$$ of $$\triangle{ABC}$$ at $$P$$ and $$AB$$,$$AC$$ produced at $$Q$$ and $$R$$ respectively.

Proof: Lengths of tangents drawn from an external point to a circle are equal.

$$\Rightarrow AQ = AR, BQ = BP, CP = CR$$.

Perimeter of $$\triangle{ABC}=AB + BC + CA$$

$$= AB +\left(BP + PC\right)+\left(AR – CR\right)$$

$$=\left(AB + BQ\right)+\left(PC\right)+\left(AQ – PC\right)$$ since $$\left[AQ = AR, BQ = BP, CP = CR\right]$$

$$= AQ + AQ$$

$$= 2AQ$$

$$\Rightarrow AQ =\dfrac{1}{2}$$(Perimeter of $$\triangle{ABC}$$)

$$\therefore AQ$$ is the half of the perimeter of $$\triangle{ABC}$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image