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$ \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} $.
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 $\angle ECD = 50^o$.