Question

Prove that the median of a triangle divides it into two triangles of equal areas.​

Answer 1

From the diagram in figure 1 we see that the two triangles CMA and BCM are equal in area. We can come up with a conjecture and say that, the median of a triangle divides the triangle into two triangles with equal areas.

To show that this is always true we can write a short proof:

Area of any triangle = half the base x height.

In the triangles CMA and CBM, AM and MB are the bases respectively. The two triangles have the same height.

Therefore area of triangle CMA= ½(AM)(FE)

And area of triangle CBM=1/2(MB) (FE)

From the two areas we see that FE=FE (the two triangles have the same height).

Also AM=MB (M is the midpoint of AB, since AM is the median of the triangle. This then means that the two triangles are equal in area.

Now let us consider two medians. Look at the diagram below. We want to see if we can say anything else about the areas.

We have already seen that the median divides the triangle in two equal areas.

Let us extend that and see what happens when we put in the second median.

Look at the GSP diagram in figure 2

Answer 2

SOLUTION:-

Given:

∆ABC is a triangle. AD is the median of ∆BCA.

Construction:

AC perpendicular to BC.

To prove:

Area of (∆BAD) = Area of (∆ADC)

Proof:

AD is the median of ∆ABC.

BD = DC [D is the mid- point of BC]

Therefore,

We can multiply both by AL.

BD × AL = DC× AL

We can multiply both by 1/2.

1/2 × BD × AL = 1/2 ×DC × AL

Thus,

Area of (∆ABD) = Area of (∆ADC)

Proved.