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 ha and hb to the sides a and b (not necessarily orthocenter). The quadrilateral AFBH is a parallelogram.


The parallelogram related to the vertex C

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 Mc of AB lies on FH and, in addition, HMc = McF.

Corollary 2: Because CO=OF and HMc = McF (according to the definition of point F and corollary 1), line OMc is a midline of ΔCHF.

Exception: If ∠ACB = 90, then points C and H (i.e. points O and Mc) are identical.

You can also see a related applet