Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

∴ Area (ΔDAC) = Area (ΔDBC)

⇒ Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)

⇒ Area (ΔAOD) = Area (ΔBOC)

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In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

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 the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

(i) ar(PRQ) = 1/2 ar(ARC)

(ii) ar(RQC) = 3/8 ar(ABC)

(iii) ar(PBQ) = ar(ARC)

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 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)

ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (ABC) = 24 cm².

In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).

In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)