Question

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

Answer 1

It is given that, D is any point on side AC of a ∆ABC with AB = AC.



In ∆ABC,

AB = AC (Given)

∴ ∠ACB = ∠ABC (In a triangle, equal sides have equal angles opposite to them)

Now, ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC (∠ACB = ∠ABC)

In ∆BCD,

∠DCB > ∠DBC

⇒ BD > CD (In a triangle, greater angle has greater side opposite to it)

Or CD < BD