MCQ
In the given figure $ABCD$ is a cyclic quadrilateral in which $AB || DC$ and $\angle\text{BAD}=100^\circ.$ Then, $\angle\text{ABC}=?$
- ✓$100^\circ$
- B$40^\circ$
- C$50^\circ$
- D$80^\circ$
✓
Answer
Correct option: A.
$100^\circ$
Since $ABCD$ is a cyclic quadrilateral, we have:
$\angle\text{BAD}+\angle\text{BCD}=180^\circ ($Opposite angles of a cyclic quadrilateral$)$
$\Rightarrow100^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=(180^\circ-100^\circ)=80^\circ$
Now, $AB || DC$ and $CB$ is the transversal.
$\therefore\angle\text{ABC}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{ABC}+80^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-80^\circ)=100^\circ$
$\Rightarrow\angle\text{ABC}=100^\circ$
$\angle\text{BAD}+\angle\text{BCD}=180^\circ ($Opposite angles of a cyclic quadrilateral$)$
$\Rightarrow100^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=(180^\circ-100^\circ)=80^\circ$
Now, $AB || DC$ and $CB$ is the transversal.
$\therefore\angle\text{ABC}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{ABC}+80^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-80^\circ)=100^\circ$
$\Rightarrow\angle\text{ABC}=100^\circ$
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