Consider an equilateral triangle ABC. Since each angle is 60° in an equilateral triangle, we have ∠A = ∠B = ∠C = 60° and the sides of equilateral triangle is AB = BC = CA = 2a units.

Draw the perpendicular line AD from vertex A to BC as shown in the given figure. Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the equilateral triangle ABC.

Therefore, ∠BAD = ∠CAD = 30°. Since point D divides the side BC into equal halves.

Consider right angle triangle ABD in the above given figure.

We have AB = and BD =

Then AD² = AB² – BD²

(By Pythagoras theorem)

= (2a)² – (a)² = a²

Therefore, AD = √3a² = √3a

BD = a, AD = √3a and hypotenuse = AB = 2a and ∠DAB = 30°.