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$ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC =70â°$, $\angle BAC=30â°$. Find $\angle BCD$. Further if $AB=BC$, find $\angle ECD$.
Given :
$ABCD$ is a cyclic quadrilateral.
$\angle DBC =70⁰$, $\angle BAC=30⁰$ and $AB=BC$.
To do :
We have to find $\angle BCD$ and $\angle ECD$.
Solution :
In the cyclic quadrilateral $ABCD$, it is given that $AB=BC$.
Therefore, $\triangle ABC$ is an isosceles triangle.
So, $\angle BAC=\angle BCA=30°$.
$DC$ is a chord, which subtends the angles $\angle DAC$ and $\angle DBC$ on the circle.
Therefore, $\angle DAC=\angle DBC$ [Angles on the same segment are equal]
$\angle DAC=\angle DBC= 70°$
$\angle BAD=\angle DAC+ \angle BAC$
$\angle BAD=70°+30°= 100°$
Opposite angles are supplementary in a cyclic quadrilateral.
$\angle BAD+\angle BCD= 180°$
$100°+\angle BCD=180°$
$\angle BCD=180°-100°$
$\angle BCD=80°$
$\angle BCD=\angle BCA+\angle ACD $
We already know that, $\angle BCA=30°$.
$80°=30°+\angle ACD$
$\angle ACD=80°-30°$
$\angle ACD=\angle ECD=50°$
Therefore, $\angle BCD=80°$ and $\angle ECD=50°$.