MathNet 2019-part-07 — 184 problems (EN–VI) Question 75

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 33