Question

ABC is an equilateral triangle and D is any point in AC. Prove that BD > AD.

Answer 1

Hint: We will first start by using the fact that $\Delta ABC$ is an equilateral triangle. Therefore, the angles of $\Delta ABC$ are $60{}^\circ $ each. Then we will prove that $\angle BAD<\angle ABD$ and will use the property of triangle that side opposite larger angle is greater than the side opposite smaller angle.

Complete step-by-step answer:
Now, we have been given that ABC is an equilateral triangle and D is any point in AC and we have to prove that BD > AD.

Now, we know that in an equilateral triangle each angle is $60{}^\circ $. Therefore, we have $\angle BAC=60{}^\circ \ and\ \angle ABC=60{}^\circ $.
Now, we have from the figure that,
$\begin{align}
  & \angle ABC=\angle ABD+\angle DBC \\
 & \angle ABD+\angle DBC=\angle ABC \\
\end{align}$
Now, we will substitute $\angle ABC=60{}^\circ $.
$\angle ABD+\angle DBC=60{}^\circ $
Or we can say that,
$\angle ABD<60{}^\circ $
Now, we will substitute $60{}^\circ =\angle BAD$. So, we have,
$\angle ABD<\angle BAD$
Now, we know that the side opposite to larger angle is greater than the side opposite to smaller angle. So, we have,
\[AD < BD\]
Hence Proved.

Note: It is important to note that we have used a fact that $\angle ABC=60{}^\circ $. Then we have splitted it as $\angle ABD+\angle DBC=60{}^\circ $ and since, both the angle $\angle ABC\ and\ \angle DBC$ are positive. Therefore, if we remove one of them, their overall sum will be less than $60{}^\circ $. Therefore, we have $\angle ABD<60{}^\circ $.