Question

1) In the figure given below, E is any point on the median AD of a \[\vartriangle ABC\]. Show that $ar(\vartriangle ABE) = ar(\vartriangle ACE)$.
2) In a \[\vartriangle ABC\], E is the midpoint of the median AD. Show that $ar(\vartriangle BED) = \dfrac{1}{4}ar(\vartriangle ABC)$.

Answer 1

Hint: 1) We will first write the theorem related to the median and the areas of the triangle to get the relation of areas of some triangle by using them. Then, we will form two equations using the same theorem and then subtract them to get the required result.
2) We will just use the above stated fact and move ahead using all of its equations.

Complete step-by-step answer:
Part 1):
Let us first of all write the theorem we will require:
The theorem states that a median divides the triangle into two triangles of equal areas.
Now, let us look at \[\vartriangle ABC\]:
We are given that AD is the median of \[\vartriangle ABC\]. Therefore, AD must divide it into two triangles of equal areas. And also, D must be the mid point of BC.
Hence, BD = DC ………..(1)
We can clearly see that when we form AD, we get triangles on both sides of the median named as \[\vartriangle ABD\] and \[\vartriangle ACD\].
Hence, $ar(\vartriangle ABD) = ar(\vartriangle ACD)$. …………..(2)
Now, let us look at \[\vartriangle EBC\]:
We are given that D is the midpoint of BC. So, ED is the median of \[\vartriangle EBC\]. (Using (1))
Therefore, ED must divide it into two triangles of equal areas.
We can clearly see that when we form ED, we get triangles on both sides of the median named as \[\vartriangle EBD\] and \[\vartriangle ECD\].
Hence, $ar(\vartriangle EBD) = ar(\vartriangle ECD)$. …………..(3)
Now, subtracting (3) from (2), we will get:-
$ \Rightarrow ar(\vartriangle ABD) - ar(\vartriangle EBD) = ar(\vartriangle ACD) - ar(\vartriangle ECD)$
We can clearly see from the diagram that it will result in:
$ \Rightarrow ar(\vartriangle ABE) = ar(\vartriangle ACE)$
Hence, proved.
Part 2):
Now, since we know that a median divides a triangle into half.
We are given that AD is the median of \[\vartriangle ABC\].
Therefore, $ar(\vartriangle ABD) = \dfrac{1}{2}ar(\vartriangle ABC)$ ………….(4)
Now, since E is the midpoint of AD. Hence, BE is the median of \[\vartriangle ABD\].
Now, using the same fact that a median divides a triangle into half.
Therefore, $ar(\vartriangle BED) = \dfrac{1}{2}ar(\vartriangle ABD)$ ………………(5)
Putting (4) into (5), we will get:-
$ \Rightarrow ar(\vartriangle BED) = \dfrac{1}{2}\left[ {\dfrac{1}{2}ar(\vartriangle ABC)} \right]$
This is equivalent to:
$ \Rightarrow ar(\vartriangle BED) = \dfrac{1}{4}ar(\vartriangle ABC)$.
Hence, proved.

Note: The students must note that they need not cram the theorem but they can actually visualize it very easily. Let us look at why this happens that a median divides a triangle into half:
We know that the area of a triangle is given by $A = \dfrac{1}{2} \times base \times height$. When we draw a median and look at two triangles, height remains the same, only the base is changed to half. Hence, it is divided into half.