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

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Given:

$\mathrm{ABCD}$ is a cyclic quadrilateral whose diagonals intersect at a point $\mathrm{E}$.

$\angle \mathrm{DBC}=70^{\circ}$, $\angle \mathrm{BAC}$ is $30^{\circ}$

$\mathrm{AB}=\mathrm{BC}$
To do:

We have to find $\angle \mathrm{BCD}$ and $\angle \mathrm{ECD}$.

Solution:

We know that,

The angles in the same segment of a circle are equal.

This implies,

$\angle BDC = \angle BAC=30^o$

In $\triangle BCD$,

$\angle BDC + \angle DBC + \angle BCD = 180^o$

$30^o+70^o+\angle BCD=180^o$

$100^o+\angle BCD=180^o$

$\angle BCD=180^o-100^o$

$\angle BCD=80^o$

$AB = BC$

This implies,

$\angle BCA = \angle BAC= 80^o$      (Angles opposite to equal sides in a triangle are equal)

$\angle ECD = \angle BCD - \angle BCA$

$= 80^o - 30^o$

$= 50^o$

Hence, $\angle BCD = 80^o$ and $\angle ECD = 50^o$.

Updated on 10-Oct-2022 13:46:42