MCQ
In the adjoining Figure, $AB = AC$ and $BO = CD$. The ratio $\angle\text{ABO} : \angle\text{ACD}$ is: 
- ✓It is $1 : 1$
- BIt is $1 : 2$
- CIt is $2 : 3$
- DIt is $2 : 1$
✓
Answer
Correct option: A.
It is $1 : 1$
In $\triangle\text{ABC}$
$\text{AB = AC}$
$\therefore \angle\text{ABC} = \angle\text{ACB}$ (angles opposite to equal sides of a triangle are equal)$ ...(1)$
In $\triangle\text{DBC},$
$\text{DB = DC},$
$\therefore \angle\text{DBC} = \angle\text{DCB}$ (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 $\triangle\text{ACD}$
$\Rightarrow\frac{\angle\text{ABD}}{\angle\text{ACD}}=1$
$\therefore \angle\text{ABD} : \angle\text{ACD}=1:1$
$\text{AB = AC}$
$\therefore \angle\text{ABC} = \angle\text{ACB}$ (angles opposite to equal sides of a triangle are equal)$ ...(1)$
In $\triangle\text{DBC},$
$\text{DB = DC},$
$\therefore \angle\text{DBC} = \angle\text{DCB}$ (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 $\triangle\text{ACD}$
$\Rightarrow\frac{\angle\text{ABD}}{\angle\text{ACD}}=1$
$\therefore \angle\text{ABD} : \angle\text{ACD}=1:1$
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