Question 22
In the adjoining figure, ABCD is a cyclic quadrilateral. If AB is a diameter, BC = CD and $$\angle ABD = 40^\circ$$, find the measure of $$\angle DBC$$.
Solution
Since AB is the diameter $$\angle\ ADB\ =\ 90^0$$
Which gives $$\angle\ DAB\ =\ 50^0$$
ABCD is cyclic quadrilateral so $$\angle\ A+\angle\ C\ =\ 180^0$$
Thus $$\angle\ DCB\ =\ 130^0$$
In$$\bigtriangleup\ $$ BCD, 2x+130 = 150
$$x\ =\ 25^0$$
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