$ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC =70⁰$, $\angle BAC=30⁰$. 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 $\angle ECD=50°$.

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Updated on: 10-Oct-2022

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