In a right-angled triangle ABC (angle C = 90 degrees) point O is the inner point of leg AC. Prove that AB> OB> CB.

This problem is easily solved if we use the theorem that in each triangle opposite the larger side there is a large side. Consider triangles ACB; AOB; and OCB, and the angles in these triangles.

Triangle AOB: <AOB = 180 – <COB, and from the triangle OCB <COB = (90 – <COB) <90, then <AOB = 180 – <COB = 180 – 90 + <COB = 90 + <COB. From all this it is clear that the angles are distributed in magnitude as follows: <AOB> <OCB> <ACB = 90.

Then from the triangle AOB: AB> OB; from the triangle ОSВ: ОВ> CB (leg opposite the acute angle).