Question:
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC).
Solution:
Draw AM ⊥ BC
Since, AD is the median of ΔABC
∴ BD = DC
⇒ BD = AM = DC × AM
⇒ (1/2)(BD × AM) = (1/2)(DC × AM)
⇒ ar(ΔABD) = ar(ΔACD) ⋅⋅⋅⋅⋅⋅ (1)
In ΔBGC, GD is the median
⇒ ar(ΔBGD) = ar(ΔCGD) ⋅⋅⋅⋅⋅⋅ (2)
In ΔACD, CG is the median
⇒ ar(ΔAGC) = ar(ΔCGD) ⋅⋅⋅⋅⋅ (3)
From (2) and (3) we have,
ar(ΔBGD) = ar(ΔAGC)
But, ar(ΔBGC) = 2ar(ΔBGD)
⇒ ar(ΔBGC) = 2ar(ΔAGC)