Question

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.

Answer 1

Given: In $∆$ABC, AB = AC

To prove: CD < BD

Proof:

In $∆$ABC,

Since, AB = AC (Given)

So, $\angle ABC=\angle ACB$ ...(i)

$$\begin{gathered} \mathrm{In}∆ABC\mathrm{and}∆DBC, \\ \angle ABC>\angle DBC \\ \Rightarrow \angle ACB>\angle DBC\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow BD>CD\left(\mathrm{Side}\mathrm{opposite}\mathrm{to}\mathrm{greater}\mathrm{angle}\mathrm{is}\mathrm{longer}.\right) \\ \therefore CD<BD \end{gathered}$$