Question

4. Diagonals of cyclic quandrilateral ABCD intersected at E. If DBE=70°,BAC=30,then find BCD and if AB=BC, then find ECD.
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Answer 1

In the triangles, ABD and BCD, ∠CAD = ∠CBD = 70°. (Angles in the same segment are equal)

Hence, ∠BAD = ∠CAB + ∠DAC

= 30° + 70° = 100°

Thus, ∠BAD = 100°

Since ABCD is a cyclic quadrilateral, the sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

∠BAD + ∠BCD = 180°

∠BCD = 180° - 100°

= 80°

Thus, ∠BCD = 80°

Also given AB = BC.

So, ∠BCA = ∠BAC = 30° (Base angles of isosceles triangle are equal)

∠ECD = ∠BCD - ∠BCA

= 80° - 30°

= 50°

Thus, ∠ECD = 50°