Question

In ∆ABC, D is a point in side BC such that 2BD = 3DC. Prove that the area of ∆ABD = 3/5 × Area of ∆ABC

Answer 1

It is given that ABC is an equilateral triangle, which means all sides are equal.

Let say the mesaurement for a side AB be 3x

i.e., AB=CA=BC= 3x

Now, it is given that BD=1/3 BC.

Therefore, BD= (1/3)*3x= x

Let’s look forward to triangle ABD

Angle ABD= 60 degree, as all the angles in an equilateral triangle measures 60 degree.

AB= 3x, BD= x

with this information let’s find out AD’s length with the help of Cosine Rule.

Cosine Rule = 2abCosC= a(square) + b (square) - c (square)

Putting the values in the rule we get,

2*3x*x*cos(60) = (3x)square + (x)square- AD(square)

6x(square)*1/2= 9(x)Square +(x)square - AD(square)

AD(square)= 10x(square) - 3x(square)

AD= (square root)7x

Now we can just put the values of sides we need to get ratio of.

AD:AB= (square root)7x : 3x

AD: AB= (square root) 7 : 3