Through the middle of the hypotenuse of the right-angled triangle ABC, the perpendicular KO is drawn to its plane.

Through the middle of the hypotenuse of the right-angled triangle ABC, the perpendicular KO is drawn to its plane. Prove that the oblique KA, KB and KC are equal. Calculate the lengths of the projections of these oblique triangles on the plane if AC = BC = a.

Since, by condition, AB = BC, the triangle ABC is rectangular and isosceles.

OK – perpendicular to the hypotenuse of the AC, then the triangles AOK and СOК are rectangular, in which the OC is common, the OA = OC by condition, since point O is the middle of the AC.

Then the triangles AOK and COK are equal in two legs, which means KA = KC.

Let’s build the height of the OB. Since ABC is isosceles, then OB is also the median of the ABC triangle, and then the angle ABO = ABC / 2 = 90/2 = 45. Then the ABO triangle is isosceles, AO = BO, and then the ВOK triangle is equal to the AOK and СOK triangles, and KB = KA = KC, as required.

The projection of the KC segment on ABC is the OC segment, the KA segment is the OA segment.

AC = a * √2 cm, then KA = KC = AC / 2 = a * √2 / 2 cm.

The ВC projection on ABC is the height BO of the triangle ABC, and we determined that BO = AO = a * √2 / 2 cm.

Answer: The projections of the oblique are equal to a * √2 / 2 cm.