Concurrency of the Altitudes (Heights) of Triangles and the Orthocentre of Triangles

In triangle \(ABC\), \(AE\) and \(BF\) are perpendicular to \(BC\) and \(AC\), respectively. \(D\) is a point on \(AB\) such that \(COD\) is a straight line.

Diagram showing triangle ABC with altitudes AE and BF intersecting at O, and point D on AB.

1. Prove the concurrency (intersecting at one point) of the \(3\) altitudes (heights) of a triangle by proving that \(CD\) is perpendicular to \(AB\).

Hint: Draw lines that are parallel to \(BC\), \(AC\) and \(AB\) through points \(A\), \(B\) and \(C\), respectively, then use the fact that the \(3\) perpendicular bisectors of the sides of a triangle intersect at one point to prove that \(CD\) is perpendicular to \(AB\).

2. Hence prove that there exists a circle with a diameter \(OB\) that passes through the \(4\) points \(O\), \(D\), \(B\) and \(E\).
3. From the diagram above, state another group of \(4\) points that lie on the same circle and state one diameter of the circle.

Note: The point \(O\) is known as the orthocentre of triangle \(ABC\) and there are totally \(6\) groups of \(4\) points in the above diagram that lie on the same circle.

Let \(O\) be the intersection of the altitudes \(AE\) and \(BF\).

The six distinct four-point groups refer to the acute, nondegenerate configuration shown; a right triangle has coincident altitude feet and does not give six distinct groups.