Question

$D$ is any point on side $AC$ of a $\Delta ABC$ with $AB=AC$. Show that $CD<BD.$

Answer 1

Consider $△ABC$

It is given that $AB=AC$

So we get

$\angle ABC=\angle ACB\left(1\right)$

From the figure, we know that

$\angle ABC=\angle ADB+\angle DBC$

So we get

$\angle ABC>\angle DBC$

From equation (1)

$\angle ACB>\angle DBC$

i.e. $\angle DCB>\angle DBC$

It means that

$BD>CD$

So we get

$CD<BD$

Therefore, it is proved that $CD<BD$.