Question

In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in D . Prove that AD square = 3 BD square

Answer 1

Hi !

In an equilateral triangle ABC ,
AB = BC = AC
AD ⟂ BC
AD is also the perpendicular bisector of the side BC.
Hence, BD = DC
Also,
BC = 2 BD

According to Pythagoras theorem,
In ΔADB,

AB² = AD² + BD²

AD² = AB² - BD²

AB = BC ,

AD² = BC² - BD²

AD² = (2BD)² - BD²

AD² = 4BD² - BD²

AD² = 3BD²

Proved.

Answer 2

Hey mate

Here is ur answer