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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