Question

Select the statements that are true for the given figure.

• A. $AP+BQ+CR=BP+CQ+AR.$
• B. $AP+BQ+CR=2(BP+CQ+AR)$
• C. $AP+BQ+CR=\frac{1}{2}\times$ Perimeter of $\Delta ABC$
• D. $AP+BQ+CR=\frac{1}{3}\times$ Perimeter of $\Delta ABC$

Answer 1

Answer: (A), (C)

The correct options are
A

$AP+BQ+CR=BP+CQ+AR.$

C

$AP+BQ+CR=\frac{1}{2}\times$ Perimeter of $\Delta ABC$

Given - In the figure, sides of $\Delta$ ABC touch a circle at P,Q,R

Consider AP+BQ+CR

$\because$ From B, BQ and BP are the tangents to the circle.

$\therefore$ BQ = BP. ....(i)

Similarly we can prove that

AP= AR ....(ii)

and CR = CQ ....(iii)

Adding we get

AP+BQ+CR=BP+CQ+AR ....(iv)

Adding AP+BQ+CR both sides

2(AP+BQ+CR)=AP+PQ+CQ+QB+AR+CP

$\Rightarrow$ 2(AP+BQ+CR)=AB+BC+CA

$\therefore$ AP+BQ+CR=$\frac{1}{2}$(AB+BC+CA)

$=\frac{1}{2}$ Perimeter of $\Delta$ ABC