Question
In Fig. $AB = AC $and $DB = DC$, find the ratio $\angle\text{ABD}:\angle\text{ACD}.$
✓
Answer
Consider the figure.
Given, $AB = AC, DB = DC$ and given to find the ratio $\angle\text{ABC}=\angle\text{ACD}$
Now, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are isosceles triangles since $\text{AB}=\text{AC}$ and $\text{DB}=\text{DC}$ respectively $\angle\text{ABC}=\angle\text{ACB}$ and $\angle\text{DBC}=\angle\text{DCB}$ [Angles opposite to equal sides are equal]
Now consider, $\angle\text{ABC}:\angle\text{ACD}$
$(\angle\text{ABC}-\angle\text{DBC}):(\angle\text{ACB}-\angle\text{DCB})$
$(\angle\text{ABC}-\angle\text{DBC}):(\angle\text{ABC}-\angle\text{DBC})$
$[\angle\text{ABC}=\angle\text{ABC}=\angle\text{ACB}\text{ and }\angle\text{DBC}=\angle\text{DBC]}$
$1:1$
$\text{ABC}:\text{ACD}=1:1$
Given, $AB = AC, DB = DC$ and given to find the ratio $\angle\text{ABC}=\angle\text{ACD}$
Now, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are isosceles triangles since $\text{AB}=\text{AC}$ and $\text{DB}=\text{DC}$ respectively $\angle\text{ABC}=\angle\text{ACB}$ and $\angle\text{DBC}=\angle\text{DCB}$ [Angles opposite to equal sides are equal]
Now consider, $\angle\text{ABC}:\angle\text{ACD}$
$(\angle\text{ABC}-\angle\text{DBC}):(\angle\text{ACB}-\angle\text{DCB})$
$(\angle\text{ABC}-\angle\text{DBC}):(\angle\text{ABC}-\angle\text{DBC})$
$[\angle\text{ABC}=\angle\text{ABC}=\angle\text{ACB}\text{ and }\angle\text{DBC}=\angle\text{DBC]}$
$1:1$
$\text{ABC}:\text{ACD}=1:1$
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