Corollary: Triangles on the same base and between the same parallels are equal in area.

In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).

In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :

(1) ar (Δ PBQ) = ar (Δ ARC)

(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)

(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .

In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).

In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that

(1)  ar (Δ BDE) = `1/2` ar (ΔABC) 

(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)

(3)  ar (BEF) = ar (ΔAFD)

(4) area (Δ ABC) = 2 area (ΔBEC)

(5) ar (ΔFED) `= 1/8` ar (ΔAFC) 

(6) ar (Δ BFE) = 2 ar (ΔEFD)

In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:

(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)

XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that

ar (ABE) = ar (ACF)

In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (figure), then prove that ar (BPQ) = `1/2` ar(∆ABC).

If the medians of a ∆ ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)