Let AB and BC be chords of a circle ( with
|AB| > |BC| ). Let M be the midpoint of
arc ABC and F the foot of the perpendicular
from M to AB.

Prove that |AF| = ( |AB| + |BC| )/2.

Let the diameter, MN, intersect the chords AB and AC at points D and E,
respectively. Let MF cross the circle again at P, and let Q be the point
on AB such that NQ is parallel to PM.

Since M is the mid-point of arc ABC, by symmetry MN is the perpendicular
bisector of AC.

The right angled triangles ADE and MDF are similar (vert. op. angles at D).

So / DAE = / DMF, and therefore the arcs subtending these angles at the

circumference are also equal, giving equal chords:  BC = PN.

       AF – BC      = AF – PN
                        = AQ
                        = BF
                        = AB – AF

Which gives:      AF = (AB + BC)/2

Posted by Harry on 2013-09-20 16:51:33