MCQ
$\text{ABCD}$ is a parallelogram in which diagonal $AC$ bisects $\angle BAD$. If $\angle BAC =35^{\circ}$, then $\angle ABC =$
- A$70^{\circ}$
- B$120^{\circ}$
- ✓$110^{\circ}$
- D$90^{\circ}$
✓
Answer
Correct option: C.
$110^{\circ}$
Given,
$\text{ABCD}$ is a parallelogram
Diagonal $\text{AC}$ bisects $\angle BAD$
$\angle B A C=35^{\circ}$
$\because \angle A +\angle B =180^{\circ} \ldots (i) [$ angle sum property of quadrilateral$]$
$\angle A =2 \angle BAC =2 \times 35^{\circ}=70^{\circ}$
Putting value of $\angle A$ in equation $(i)$
$70^{\circ}+\angle B=180^{\circ}$
$\angle B=180^{\circ}-70^{\circ}=110^{\circ}$
$\angle A B C=110^{\circ}$
$\text{ABCD}$ is a parallelogram
Diagonal $\text{AC}$ bisects $\angle BAD$
$\angle B A C=35^{\circ}$
$\because \angle A +\angle B =180^{\circ} \ldots (i) [$ angle sum property of quadrilateral$]$
$\angle A =2 \angle BAC =2 \times 35^{\circ}=70^{\circ}$
Putting value of $\angle A$ in equation $(i)$
$70^{\circ}+\angle B=180^{\circ}$
$\angle B=180^{\circ}-70^{\circ}=110^{\circ}$
$\angle A B C=110^{\circ}$
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