In the given figure, D is the mid-point of BC and E is any point on AD. Prove that :

(i) ar (△EBD) = ar (△EDC)

(ii) ar (△ABE) = ar (△ACE)

(i) Given,

BD = DC

Thus, ED is the median of triangle EBC.

We know that,

Median ED divides a triangle EBC into two triangles EBD and ECD of equal area.

∴ ar (△EBD) = ar (△EDC)

Hence, proved that ar (△EBD) = ar (△EDC).

(ii) Given,

BD = DC

Thus, AD is the median of triangle ABC.

Median AD divides a triangle ABC into two triangles ABD and ADC of equal area.

ar (△ABD) = ar (△ADC)

ar (△ABE) + ar (△EDB) = ar (△ACE) + ar (△EDC)

ar (△ABE) + ar (△EDB) = ar (△ACE) + ar (△EDB) [ar (△EBD) = ar (△EDC)]

ar (△ABE) + ar (△EDB) - ar (△EDB) = ar (△ACE)

ar (△ABE) = ar (△ACE).

Hence, proved that ar (△ABE) = ar (△ACE).