2001 HKCEE Maths MC Q46

Angle BAD = Angle DCE = 75* (opp. angles, conc. quad.)

Angle BCD = 180* - 75* = 105* (adj. angles on st. line)

Now, let angle ABC = x

Draw a line joining AC.

Since arc AB = arc BC, hence AB = BC

Then triangle ABC is an isosceles triangle

Hence, angle BAC = angle ACB = (180* - x) / 2 = 90* - x/2

Then, we have angle CAD = 75* - (90* - x/2) = x/2 - 15*

Now, in a quadrilateral, the sum of interior angles = 360*, with concylic quadrilateral, the sum of opposite angles = 180*

Hence, angle ADC + angle ABC = 180*

angle ADC = 180* - x

Since arc AB = arc BC = 1/2 arc CD

we have arc AB = arc CD

So, angle ADC = angle CAD (same arc, same angle at circumference)

So, 180* - x = x/2 - 15*

195* = 3x/2

x = 130*

So, angle ABC = 130*

The answer is D.