*Vladimir Nikolin*

**Vertex And Its Antipode - A parallelogram related to a triangle vertex **

For the definiteness sake, let's focus on vertex C and its antipode F, so that CF is a diameter of the circumcircle of the triangle ABC.

Let H be the point of intersection of the two altitudes h_{a} and h_{b} to the sides a and b (not necessarily orthocenter).
*The quadrilateral AFBH is a parallelogram.*

**Proof:** Since CF is a diameter, ∠CBF = 90° (as an inscribed angle subtended by a diameter, implying BF \perp BC. Also, AH being an altitude in ΔABC, is perpendicular to BC: AH \perp BC. Therefore, AH||BF.

Similarly, BH is an altitude (BH \perp AC) while ∠CAF = 90°. Therefore AF||BH.

**Corollary 1:**
The diagonals AB and FH of the parrallelogram AFBH cross at their common midpoint which means that the midpoint M_{c} of AB lies on FH and, in addition, **HM _{c} = M_{c}F**.

**Corollary 2:** Because CO=OF and HM_{c} = M_{c}F (according to the definition of point F and corollary 1), **line OM _{c} is a midline of ΔCHF**.

Exception: If ∠ACB = 90°, then points C and H (i.e. points O and M_{c}) are identical.

You can also see a related applet