Question
$D$ is any point on the side $AC$ of $\triangle\text{ABC}$ with $AB = AC$. Show that $CD < BD$.
✓
Answer
Given: In $\triangle\text{ABC, AB = AC}$
To prove: $\text{CD < BD}$ proof: In $\triangle\text{ABC},$
Since, $\text{AB = AC}$ (Given)
so, $\angle\text{ABC}=\angle\text{ACB}...(\text{i})$ In
$\triangle\text{ABC}$ and $\triangle\text{DBC},$
$\angle\text{ABC}>\angle\text{DBC}$
$\Rightarrow\text{ACB}>\angle\text{DBC}$ $[$from $(i)]$
$\Rightarrow\text{BD > CD}$ (Side opposite to greater angle is longer.)
$\therefore\text{CD < BD}$
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