Question

Let the vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that angle ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer 1

see the diagram.

ΔAOD and ΔCOE are Isosceles and Congruent.  Reasons:
     AD = CE  given,      AO = OD = OC = OE = Radius.
So  ∠DAO = ∠ADO = ∠ECO = ∠CEO = a (say)

Let ∠AOC = x,   ∠DOE = y   and  ∠ABC = z.    To prove that 2 z = | x - y |

At O,  ∠x + ∠y  = 2π - ∠AOD - ∠COE
                        = 2π - 2 (π - 2a)
           ∠x + ∠y = 4 a    ----- (1)

In the Isosceles ΔODE,   ∠ODE = ∠OED = (π-y)/2

At point D or at E,    ∠BED =∠BDE = π - a - (π-y)/2 = π/2 - a + y/2

In ΔBDE,  z = π - 2 * [π/2 - a + y/2 ]
         or,      ∠ABC = 2 a - y = (x+y)/2 - y/2
                              = (x - y)/2