Question

10.ABCD is a cyclic quadrilateral and AC is a diameter of

the larger circle with centre O. The smaller circle through
O, B and E meets BD at E.
Prove that:
a) Ê, = 2 × Â₁
b) AEDC is isosceles
C) OE || AD.

Answer 1

See proofs below

Explanation:

(a) If we let angle A1 = α, then angle O1 = 2α, since the angle at the centre is twice the angle at the circumference (both angles are subtended by the arc/chord BC). And therefore angle E1 also equals 2α, because angles in the same segment are equal (also subtended by arc/chord BC).

(b) Angle DEC = 180 - 2α since angles on a straight line sum to 180 degrees. Angle D1 is equal to α since angles in the same segment (once again BC) are equal. Therefore angle C1 = α since angles in a triangle sum to 180 degrees. Now, as angles C1 and D1 are equal, triangle EDC is isosceles since there are two equal base angles.