Question

a circle is touching the side QR of triangle PQR at point M and touching PQ and PR produced at S and T respectively prove that PS= 1/2 perimeter of triangle PQR​

Answer 1

Answer:

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Answer 2

Answer:

Here we  proved that PS = $\frac{1}{2}$ (perimeter of the ΔPQR )

Step-by-step explanation:

Explanation:

Given , PQR is a triangle in which a circle touching the side QR and touching PQ and PR produced at S and T.

Let x be the length of PS

Step1:

Now , PS = PT        (tangent from same external points are equal)

So, PS + PT = x + x = 2x  

⇒ PQ+QS +PR+RT = 2x     (where PS = PQ+QS and PT = PR+RT)

But  we have , QS = QM and RT = RM

Therefore ,

Tangent from same external points are equal

⇒ PQ+QS +PR+RT = 2x

⇒ \frac{(PQ+QM+RM+PR)}{2}= 2x

⇒x = $\frac{(PQ+QM+RM+PR)}{2}$  

But , here we can see that   sum of (PQ+QM+RM+PR) is equal to perimeter of the triangle  and 'x' = PS

Final answer :

Hence , we proved that PS = $\frac{1}{2}$ (perimeter of the ΔPQR )

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