Question

Q is a point on the side SR of triangle PSR such that PR . prove That PS > PR

Answer 1

$<b>In ∆PQR$

$\textbf{PQ = PR (given)}$

$\bf\Longrightarrow{PQR = ∠PRQ}...$...(Angles opposite to equal sides are equal)

∠PQR is an exterior angle of ∆PSQ

$\bf{∴ ∠PQR > ∠PSQ}$

$\bf{∠PRQ > ∠PSQ........(from 1)}$

∴In ∆PSR,

$\bf{∠PRS > ∠PSR}$

$\bf\bold\Longrightarrow{PS > PR}$ (side opposite to greater side is greater)

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