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State true or false:
A circle touches the side BC of ΔABC at P and touches AB and AC produced at Q and R respectively. Does $AQ = \frac{1}{2} (Perimeter of (ΔABC)$

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True

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answer is A.

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Detailed Solution

AQ = AR . . . . . (1)

BP = BQ . . . . . (2)

CR = CP . . . (3)

Thus, Perimeter of triangle
⇒

ABC = AB + BC + AC

From the diagram, we can see that
⇒

BC = BP + PC

and

AC = AR - CR

So, Perimeter of the triangle
⇒

ABC = AB + BP + PC + AR - CR

So, Perimeter of the triangle
⇒

ABC = AB + BQ + PC + AQ - PC

⇒

AB + BQ

is equal to

AQ

Therefore, Perimeter of the triangle
⇒

ABC = AQ + AQ = 2 AQ

Now, half of the perimeter of the triangle

= \frac{1}{2} \times AQ

.
Hence, proved.

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