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

Hint: We will first start by using the fact that ΔABC is an equilateral triangle. Therefore, the angles of ΔABC are 60∘ each. Then we will prove that ∠BAD<∠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∘. Therefore, we have ∠BAC=60∘ and ∠ABC=60∘. Now, we have from the figure that, ∠ABC=∠ABD+∠DBC∠ABD+∠DBC=∠ABC Now, we will substitute ∠ABC=60∘. ∠ABD+∠DBC=60∘ Or we can say that, ∠ABD<60∘ Now, we will substitute 60∘=∠BAD. So, we have, ∠ABD<∠BAD Now, we know that the side opposite to larger angle is greater than the side opposite to smaller angle. So, we have, AD