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  • S is any point on side QR of a PQR. Show that: PQ + QR + RP > 2 PS.

S is any point on the side QR of a \trianglePQR. Show that: PQ + QR + RP > 2 PS.

Answers (1)

Given, \trianglePQR, S is any point on QR.

To prove: PQ + QR + RP > 2PS

Proof:-  we know that the sum of two sides of a triangle is greater than the third side.

In \trianglePQS,

 PQ + QS > PS

and in \trianglePSR,

PR + SR > PS

In addition,

PQ + QS + SR + PR > PS + PS

\because QS + SR = QR

\Rightarrow PQ + QR + PR > 2PS

Hence proved.

Posted by

infoexpert24

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