Question

Show that in a right angled triangle, hypotenuse is larger than any of the remaining sides.

Answer 1

Suppose ABC is a right angled triangle with $\angle B={90}^o$. Then AC is the hypotenuse. Observe that $\angle BAC<\angle B$ and $\angle BCA<\angle B$. Now the side opposite to $\angle BAC$ is BC and side opposite $\angle BCA$ is AB. Hence by proposition 2, BC < AC and AB < AC