59th Ukrainian National Mathematical Olympiad

Question

Given a right triangle ABC, let M be the midpoint of its hypotenuse AB. A perpendicular bisector of AB intersects side BC in point K. A perpendicular from K to CM intersects the ray AC in point P, which happens not to lie on the segment AC. Lines CM and BP intersect in point T. Prove that AC = TB. (Danylo Khilko) FIGURE_0 Fig.33

Accepted Answer

From KP CM it follows that CPK = MBC = ABC (fig. 33), hence the quadrilateral PAKB is cyclic. Therefore, PK is a bisector in the triangle CPT. Clearly, PK is also a height and a median of this triangle. Therefore, CPT is isosceles. In particular, it implies that PK is a perpendicular bisector of CT. Further, CK = KT, and PCK = PTK, meaning that PTK = 90^. Finally, using that KB = KA, we get KTB = KCA, implying TB = AC.