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# 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.