Question

If E is the mid point of median AD in triangle ABC, then prove that:
Area($\vartriangle $ABE) = $\dfrac{1}{4}$ Area($\vartriangle $ABC).

Answer 1

Hint: First of we will make the diagram of the question. The median AD will divides the $\vartriangle $ABC into two triangles of equal area. And also the E is the median of the $\vartriangle $ABD. So it will also divide the $\vartriangle $ABD into two equal areas.

Complete step-by-step answer:
The diagram for the question is shown below:

AD is the median of $\vartriangle $ABC. And BE is the median of $\vartriangle $ABD.
We know that a median divides the triangle into two triangles whose areas are equal.
$\because $ AD is the median of $\vartriangle $ABC.
$\therefore $ Using this theorem, we can write:
Area($\vartriangle $ABD) = Area($\vartriangle $ADC). ---- (1)
Similarly, since BE is the median of $\vartriangle $ABD.
$\therefore $ Using the above theorem, we can write:
Area($\vartriangle $ABE) = Area($\vartriangle $BDE). ---- (2)
But, Area($\vartriangle $ABC) = Area($\vartriangle $ABD) +Area($\vartriangle $ACD) --- (3)
Also, Area($\vartriangle $ABD) = Area($\vartriangle $ABE) +Area($\vartriangle $BDE) ----(4)
Putting the values from equations 1 and 4 into equation 3, we get:
Area($\vartriangle $ABC) = 2 ( Area($\vartriangle $ABE) +Area($\vartriangle $BDE) ) .
Using equation 2, we can write:
  Area($\vartriangle $ABC) = 4 ( Area($\vartriangle $ABE) ) .
$ \Rightarrow $ Area($\vartriangle $ABE) = $\dfrac{1}{4}$ Area($\vartriangle $ABC).

Note: In the question involving the area of the triangle, you have to remember two theorems mainly. First one is that the median of a triangle divides the triangle into two triangles having equal area. Second is that triangles on the same base and between parallel lines have equal areas.