Question

prove that a median of a triangle divides it into two triangles of equal areas.

Answer 1

We have to first draw an altitude (say AE) perpendicular to the base BC.
Now,
We know that a median bisects the base.
Therefore,
BD=DC
Area of triangle ABD= 1/2×AE×BD------(1)
Area of triangle ACD= 1/2×AE×DC
But BD=DC
So,
Area of triangle ACD= 1/2×AE×BD-----(2)
Comparing (1) and (2)
Area of triangle ABD= Area of triangle ACD
A median of a triangle divides it into two triangles of equal areas.
Hope it helps! Please mark it as 'Brainliest'!

Answer 2

Given : In ΔABC, AD is the median of the triangle.

To prove : ar ΔABD = ar ΔADC

Construction : Draw AP ⊥ BC.

Proof : ar ΔABC = $\sf \frac{1}{2}$ × BC × AP...... (i)

ar ΔABD = $\sf \frac{1}{2}$ × BD × AP

ar ΔABD = $\sf \frac{1}{2} \times \frac{BC}{2} \times AP$

[AD is the median of ΔABC]

ar ΔABD = $\sf \frac{1}{2}$ × ar ΔABC.. (ii)

ar ΔADC = $\sf \frac{1}{2}$ × DC × AP

ar ΔADC = $\sf \frac{1}{2} \times \frac{BC}{2} \times AP$

ar ΔADC =$\sf \frac{1}{2}$ × ar ΔABC.. (iii)

From equation (i), (ii) and (iii)

ar ΔABD = ar ΔADC = $\sf \frac{1}{2}$ ar ΔABC

Hence, it is proved.