Question

show that the median of a triangle divides into two triangles of equal areas

Answer 1

in triangle abc ,AD is the median
here,BD=DC
and AE perpendicular to BC
area of triangle ABD = half BC×AE
and half DC× AE
since BD= DC
area triangle abc =area of triangle adc
hence proved

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.