MCQ
In the adjoining Figure, $AB = AC$ and $BD = CD$. The ratio $\angle\text{ABD} : \angle\text{ACD}$ is:
- A$2 : 3$
- B$2 : 1$
- C$1 : 2$
- ✓$1 : 1$
✓
Answer
Correct option: D.
$1 : 1$
In $\triangle\text{ABC}$
$AB = AC$
$\therefore\ \angle\text{ABC} = \angle\text{ACB}$ (angles opposite to equal sides of a triangle are equal)$ ...(1)$
In $\triangle\text{DBC}$
$DB = DC,$
$\therefore\ \angle\text{DBC} = \angle\text{DCB}$ (angles opposite to equal sides of a triangle are equal)$ ...(2)$
Subtract $2$ from $1$
$\angle\text{ABC} - \angle\text{DBC} = \angle\text{ACB} - \angle\text{DCB}$ (equals subtracted from equals gives equal)
$= \angle\text{ABD} = \angle\text{ACD}$
Divide both the sides by $\angle\text{ACD}$
$\Rightarrow \frac{\angle\text{ABD}}{\angle\text{ACD}} = 1$
$\therefore\ \angle\text{ABD} : \angle\text{ACD}=1:1$
$AB = AC$
$\therefore\ \angle\text{ABC} = \angle\text{ACB}$ (angles opposite to equal sides of a triangle are equal)$ ...(1)$
In $\triangle\text{DBC}$
$DB = DC,$
$\therefore\ \angle\text{DBC} = \angle\text{DCB}$ (angles opposite to equal sides of a triangle are equal)$ ...(2)$
Subtract $2$ from $1$
$\angle\text{ABC} - \angle\text{DBC} = \angle\text{ACB} - \angle\text{DCB}$ (equals subtracted from equals gives equal)
$= \angle\text{ABD} = \angle\text{ACD}$
Divide both the sides by $\angle\text{ACD}$
$\Rightarrow \frac{\angle\text{ABD}}{\angle\text{ACD}} = 1$
$\therefore\ \angle\text{ABD} : \angle\text{ACD}=1:1$
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