Question

Two disjoint circles C₁ and C₂ with centers O₁ and O₂ are given. A common exterior tangent touches circles C₁ and C₂ at A and B respectively and O₁O₂ intersects circles C₁ and C₂ at points C and D respectively. Prove that: 

(a) the points A, B, C and D are concyclic 

(b) the straight lines AC and BD are perpendicular.

Answer 1

(a) Letting the base angles in isosceles triangles AO₁C and BO₂D be x and y, respectively, the sum of the angles in quadrilateral ABDC is (90° – x) + (90° – y) + (180° – y) + (180° – x) = 360°, and we have 

x + y = 90°. .... (1) 

Hence, in ABDC, the angles at A and D add up to (90° – x) + (180° – y) = 270° – (x + y) = 270° – 90° = 180° and thus, 

ABDC is cyclic. This proves (a).

(b) Let AC and BD when produced intersect at E. It follows from equation (1) that in triangle CED the angles at C and D add up to 90°. Thus, CED is a right-angled triangle with the right angle at E and AC and BD are in fact perpendicular.