Let B and C be points on two given non-collinear rays from the same point A such that |AB| + |AC| is constant.

Prove that there exists a point D, distinct from A, such that the circumcircles of triangles ABC pass through D for all choices of B and C.

Let |AB| = k + t and |AC| = k - t, so that |AB| + |AC| = 2k (constant) and t is a parameter.
Let b and c be unit vectors in the directions of AB and AC, and let p denote the vector AP where P is the centre of the circle through A, B and C.

Perpendiculars from P to AB and AC will meet those chords at their mid points, So:
            [p - 0.5(k + t)b].b = 0    which gives        p.b = (k + t)/2               (1)

and       [p - 0.5(k - t)c].c = 0      which gives        p.c = (k - t)/2                (2)

Adding (1) and (2) gives             p.(b + c) = k

Since b + c is a vector parallel to the bisector of angle A, it follows that P lies on a line perpendicular to the bisector of angle A. If D is the reflection of A in this perpendicular then AD is a chord of the circle, proving that D lies on the circle. Since the result is independent of t, this property holds for all chosen positions of B and C.

Since b + c is a vector of length 2*cos(A/2), D lies at a distance of k*sec(A/2) from A along the bisector of angle A. Thus when t = 0, triangles ABD and ACD are congruent, with right angles at B and C.

Posted by Harry on 2010-12-19 23:17:42