Question

AD is a bisector of angle A of triangle ABC . Show that AB>BD

Answer 1

considering an equilateral triangle ABC 

with sides equal sides AB ; AC and BC 

since AD is the bisector of angle A and D is located between B & C 

therefore BC = CD + DB 

and CD = DB 

BC = 2DB 

considering the pythagorean theorem that states; 

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle). 

given by the equation; a^2 + b^2 = c^2 

considering AB as the hypothenuse, AD is the altitude or one of the leg of the triangle, DB one of the leg. 

Base from the pythagorean theorem alone AB is the side opposite of the right angle and therefore is the hypothenuse 

AB^2 = AD^2 + DB^2 

AB = sqr root (AD^2 + DB^2) 

therefore AB is greater than DB since the hypothenuse of the triangle is the longest sid

Answer 2

answer is in the attachment...

hope it helps...