ABC is the triangle and M, N are midpoints of AB and AC. Points W, X are on AB, Y, Z are on AC such that WM = MX, ZN = NY. Let T be the intersection of WY and XZ, prove that T lies on the nine point circle of ABC if WY ⊥ XZ.
Let $ \angle BAC = \alpha$.
Hint: Recall that the nine-point circle passes through the midpoint of all sides.
Step 1. To show that $T$ lies on the circle, it suffices to show that $\angle MTN = 180^\circ - \alpha$.
Step 2. To show that $\angle MTN = 180^\circ - \alpha$, we will show that $\angle MTW + \angle NTZ = 90 ^\circ- \alpha$.
Step 3. $XTW$ (resp $YTZ$) is a right angle triangle, so $XMT$ (resp $TNZ$) is an isosceles triangle.
Hence $90^\circ - \alpha = \angle TWM + \angle TZY = \angle WTM + \angle ZTN$, and we are done.