Show that in a right angled triangle, the hypotenuse is the longest side.

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.