ABCD is a cyclic quadrilateral whose diagonal | vedclass

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Question

(A) Since angles in the same segment of a circle are equal,we have $\angle BAC = \angle BDC$.Given $\angle BAC = 30^{\circ}$,therefore $\angle BDC = 3

Vedclass · Accepted Answer

$80^{\circ}; 50^{\circ}$

Vedclass · Answer

(A) Since angles in the same segment of a circle are equal,we have $\angle BAC = \angle BDC$.Given $\angle BAC = 30^{\circ}$,therefore $\angle BDC = 30^{\circ}$.In $\Delta BCD$,the sum of angles is $180^{\circ}$:$\angle BCD + \angle DBC + \angle BDC = 180^{\circ}$$\angle BCD + 70^{\circ} + 30^{\circ} = 180^{\circ}$$\angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ}$.Now,in $\Delta ABC$,since $AB = BC$,the angles opposite to equal sides are equal,so $\angle BCA = \angle BAC = 30^{\circ}$.Since $\angle BCD = \angle BCA + \angle ECD$,we have:$80^{\circ} = 30^{\circ} + \angle ECD$$\angle ECD = 80^{\circ} - 30^{\circ} = 50^{\circ}$.

$ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC = 70^{\circ}$ and $\angle BAC = 30^{\circ}$,find $\angle BCD$. Further,if $AB = BC$,find $\angle ECD$.

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