Right triangle then with equality only in the isosceles case. If segments of lengths "p" and "q" emanating from vertex "C" trisect the hypotenuse into segments of length "c"/3, then The right triangle is the only triangle having two, rather than one or three, distinct inscribed squares. Let "h" and "k" ("h" > "k") be the sides of the two inscribed squares in a right triangle with hypotenuse "c". Then These sides and the incircle radius "r" are related by a similar formula: The perimeter of a right triangle equals the sum of the radii of the incircle and the three excircles:
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