Right triangle and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse. In any right triangle the diameter of the incircle is less than half the hypotenuse, and more strongly it is less than or equal to the hypotenuse times formula_44 In a right triangle with legs "a", "b" and hypotenuse "c", with equality only in the isosceles case. If the altitude from the hypotenuse is denoted "h",
Triangle inequality angles satisfy: Therefore, and so, in particular, That means side opposite angle is shorter than side opposite the larger angle . But . Hence: A similar construction shows , establishing the theorem. An alternative proof (also based upon the triangle postulate) proceeds by considering three positions for point : (i) as depicted (which is to be proven), or (ii) coincident with (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle , which would violate the triangle postulate), or lastly, (iii) interior to the right triangle between points and (in which case angle