In the figure of question 14, if E is the mid point of median AD, then prove that :

Area (△ ABE) = $\dfrac{1}{4}$ Area (△ ABC).

We know that,

Median of a triangle divides it into two triangles of equal area.

From figure,

AD is the median of Δ ABC, so it will divide Δ ABC into two equal triangles.

∴ Area of Δ ABD = Area of Δ ADC = $\dfrac{1}{2}$ Area of Δ ABC

⇒ Area of Δ ABD = $\dfrac{1}{2}$ Area of Δ ABC ………(1)

In Δ ABD,

Since, E is mid-point of AD,

∴ BE is the median.

∴ BE will divide Δ ABD into two equal triangles.

∴ Area of Δ ABE = Area of Δ BED = $\dfrac{1}{2}$ Area of Δ ABD

⇒ Area of Δ ABE = $\dfrac{1}{2}$ Area of Δ ABD ………(2)

Substituting value of Area of Δ ABD from equation (1) in (2), we get :

⇒ Area of Δ ABE = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of Δ ABC

⇒ Area of Δ ABE = $\dfrac{1}{4}$ Area of Δ ABC.

Hence, proved that area of Δ ABE = $\dfrac{1}{4}$ area of Δ ABC.