Let AB intersect at E, and the center of the circle is O, connecting AD and DE. AD is the axis of symmetry, so d is the midpoint of BC and o is the midpoint of AD. BC=6, BD=3, AB=5, Pythagorean theorem AD=4 in ABD, circle radius OA=2, secant theorem BE*BA=BD? , BE*5=3*3, BE= 1.8, pythagorean theorem in BDE DE=2.4, AE=AB-BE=5- 1.8=3.2, OA=OE, ∠ doe = 2 ∠ bad. -triangle ADE area -2*3. 14OA? ∠DOE/360 =3. 14*2？ (1-2 * 74/360)-AE * de/2 = 7.396-3.2 * 2.4/2 = 3.56, choose a.