Question

Two circles ABCD and ABEF intersect at point A and B. If CBE and DAF are straight lines, prove that CD is parallel to EF.

Answer 1

Given:
Two circles ABCD and ABEF intersect at points A and B.
Lines CBE and DAF are straight.

To prove CD is parallel to EF.

Join AB (common chord to both circles).

Since ABCD lies on one circle, quadrilateral ABCD is cyclic.

Similarly, quadrilateral ABEF is cyclic.

In cyclic quadrilateral ABCD,
∠BAF = ∠DCB (Exterior angle = interior opposite angle)   ...[1]

In cyclic quadrilateral ABEF,

∠BAF + ∠BEF = 180° (Sum of opposite angles of cyclic  ...[2] quadrilateral)

From [1] and [2],

∠DCB + ∠BEF = 180°

Since CBE and DAF are straight lines,

Angles ∠DCB and ∠BEF are co-interior angles.

Co-interior angles on the same side of the transversal sum to 180°, implying the lines are parallel.

Therefore, CD ∥ EF.