Question
$A B C D$ is a cyclic quadrilateral whose diagonals intersect at a point $E$ . If $\angle D B C=70^{\circ}, \angle BAC$ is $30^{\circ} \quad$, find $\angle BCD$ . Further, if $A B=B C$, find $\angle BCD$. If $A B=B C$, find $\angle ECD$.
✓
Answer
$\angle CDB =\angle BAC \mid$ Angles in the same segment of a circle are equal
$= 30 ^o ....... (1)$
$\angle DBC = 70^o ...... (2)$
In $\triangle BCD$
$\angle B C D+\angle D B C+\angle C D B=180^{\circ}\left[\text { Sum of all the angles of a triangle is } 180^{\circ}\right]$
$\Rightarrow \angle B C D+70^{\circ}+30^{\circ}=180^{\circ}[\text { Using (1) and (2)] }$
$\Rightarrow \angle B C D+1000=180^{\circ}$
$\Rightarrow \angle B C D=1800-100^{\circ}$
$\Rightarrow \angle B C D=800 \ldots \ldots \ldots$
$\text { In } \triangle A B C$
$A B=B C$
$\therefore \angle B C D=\angle B A C$ [ Angles opposite to equal sides of a triangle are equal]
$=30^{\circ} \ldots \ldots . .(4)[\because \angle BAC=30 \text { (given) }]$
$\text { Now } \angle BCD=80^{\circ} \text { |From (3) }$
$\Rightarrow \angle BCA+\angle ECD=80^{\circ}$
$\Rightarrow 30^{\circ}+\angle ECD=80^{\circ}$
$\Rightarrow \angle ECD=80^{\circ}-30^{\circ}$
$\Rightarrow \angle ECD=50^{\circ}$
$= 30 ^o ....... (1)$
$\angle DBC = 70^o ...... (2)$
In $\triangle BCD$
$\angle B C D+\angle D B C+\angle C D B=180^{\circ}\left[\text { Sum of all the angles of a triangle is } 180^{\circ}\right]$
$\Rightarrow \angle B C D+70^{\circ}+30^{\circ}=180^{\circ}[\text { Using (1) and (2)] }$
$\Rightarrow \angle B C D+1000=180^{\circ}$
$\Rightarrow \angle B C D=1800-100^{\circ}$
$\Rightarrow \angle B C D=800 \ldots \ldots \ldots$
$\text { In } \triangle A B C$
$A B=B C$
$\therefore \angle B C D=\angle B A C$ [ Angles opposite to equal sides of a triangle are equal]
$=30^{\circ} \ldots \ldots . .(4)[\because \angle BAC=30 \text { (given) }]$
$\text { Now } \angle BCD=80^{\circ} \text { |From (3) }$
$\Rightarrow \angle BCA+\angle ECD=80^{\circ}$
$\Rightarrow 30^{\circ}+\angle ECD=80^{\circ}$
$\Rightarrow \angle ECD=80^{\circ}-30^{\circ}$
$\Rightarrow \angle ECD=50^{\circ}$
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