On the sides of the corner with apex to point B, points A and C were marked so that AB = BC.

On the sides of the corner with apex to point B, points A and C were marked so that AB = BC. Through points A and C were drawn straight lines perpendicular to the sides BA and BC, respectively, which intersect at point O. Prove that ray BO is the bisector of angle ABC.

Let the angle with the apex at point B be given, on the sides of which points A and C are marked so that AB = BC. Through points A and C, straight lines were drawn, perpendicular to the sides BA and BC, respectively, which intersect at point O, that is, OA⊥AB; OC⊥DC. We got two right-angled triangles BAO and BCO, which are equal in hypotenuse and leg, since 1) the BO side is common; 2) AB = BC – by construction. From the equality of the triangles, it follows that the corresponding angles are equal, that is, ∠ABO = ∠CBO, then the BO ray divides the ABC angle in half, which means that the BO ray is the bisector of the ABC angle. Q.E.D.