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Answer :

Let ABC be a right angled triangle, such that \(\angle{ABC}\) = \(90^\circ\).

We know that,

\(\therefore \) \(\angle{ABC}\) + \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\)

...(Since, the sums of all angles of a triangle are \(180^\circ\))

\(\Rightarrow \)\(90^\circ\) + \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{BCA}\) + \(\angle{CAB}\) = \(180^\circ\) - \(90^\circ\)

\(\Rightarrow \) \(\angle{BCA}\) + \(\angle{CAB}\) = \(90^\circ\)

Thus, we can say that, both the angles \(\angle{BCA}\) and \(\angle{CAB}\) are acute.

\(\therefore \) \(\angle{BCA}\) < \(90^\circ\)and

\(\therefore \) \(\angle{CAB}\) < \(90^\circ\)

\(\therefore \) \(\angle{BCA}\) < \(\angle{ABC}\) and \(\angle{CAB}\) < \(\angle{ABC}\)

Therefore, we can say,

AB < AC and BC < AC ...(Since, side opposite to greater angle is longer)

Hence, the hypotenuse AC is the longest side is proved.

- 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 \(\angle{B}\) > \(\angle{D}\).
- In figure, PR > PQ and PS bisects \(\angle{QPR}\). Prove that \(\angle{PSR}\) > \(\angle{PSQ}\).
- 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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