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Answer :

In \(\triangle{PQR}\), we have,

PR > PQ ...(given)

\(\therefore \) \(\angle{PQR}\) > \(\angle{PRQ}\) ...(i)

(\(\because \) angle opposite to longer side is greater)

We can also say,

\(\angle{a}\) = \(\angle{b}\) ...(ii)

(\(\because \) PS bisects \(\angle{QPR}\))

On adding Equation (i) and (ii), we get,

\(\angle{PQR}\) + \(\angle{a}\)> \(\angle{PRQ}\) + \(\angle{b}\) ...(iii)

Also, \(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(180^\circ\) ...(iv)

and \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\) = \(180^\circ\) ...(v)

(\(\because \) the sums of all angles of a triangle are \(180^\circ\))

From Equation (iv) and (v), we get,

\(\angle{PQS}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRS}\) + \(\angle{PSR}\) + \(\angle{b}\)

\(\because \) \(\angle{PQS}\) = \(\angle{PQR}\) and \(\angle{PRS}\) = \(\angle{PRQ}\),

we get,

\(\angle{PQR}\) + \(\angle{PSQ}\) + \(\angle{a}\) = \(\angle{PRQ}\) + \(\angle{PSR}\) + \(\angle{b}\)

Rearranging the terms, we get,

\(\angle{PQR}\) + \(\angle{a}\) + \(\angle{PSQ}\) = \(\angle{PRQ}\) + \(\angle{b}\) + \(\angle{PSR}\) ...(vi)

From Equation (iii) and (vi), we get,

\(\angle{PSQ}\) < \(\angle{PSR}\) ...

(\(\because \) side opposite to greater angle is longer)

Hence, proved.

- Show that in a right angled triangle, the hypotenuse is the longest side.
- In figure, sides AB and AC of are extended to points P and Q respectively. Also, \(\angle{PBC}\) AB.
- In figure, \(\angle{B}\)
- AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that \(\angle{A}\) > \(\angle{C}\) and \(\angle{B}\) > \(\angle{D}\).
- Show that of all line segments drawn from a give point not on it, the perpendicular line segment is the shortest.

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