P is a point in the interior of a parallelogram ABCD. Show that

`ar (Delta APD) + ar ( DeltaPBC) = ar ( DeltaAPB) + ar (DeltaPCD)`

P is a point in the interior of a parallelogram ABCD. Show that (i) ar (DeltaAPB) +ar (DeltaPCD)=1/2ar (ABCD) (ii) ar(DeltaAPD)+ar (DeltaPBC)=ar(DeltaAPB)+ar(DeltaPCD) (Hint : Throught , P draw a line parallel to AB)

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD show that ar(DeltaAPB) = ar Delta(BQC)

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (DeltaACB) = ar (DeltaACF) (ii) ar (AEDF) = ar (ABCDE)

In a triangle ABC (see figure), E is the midpoint of median AD, show that (i) ar DeltaABE = ar DeltaACE (ii) ar DeltaABE=1/4ar(DeltaABC)

DeltaABC~DeltaPQR , AB:PQ=3:4 then ar DeltaABC : ar DeltaPQR =……

In the figure , AP || BQ || CR. Prove that ar (DeltaAQC) = ar (DeltaPBR) .

From the given figure, ar ( DeltaADE ): ar ( DeltaABC )= …………..

In the figure D, E are points on the sides AB and AC respectively of DeltaABC such that ar(DeltaDBC) = ar(DeltaEBC) . Prove that DE || BC.

Draw two triangles ABC and DBC on the same base and between the same parallels as shown in the figure with P as the point of intersection of AC and BD. Draw CE || BA and BF || CD such that E and F lie on line AD. Can you show ar(DeltaPAB) = ar(DeltaPDC) Hint : These triangles are not congruent but have equal areas.

ABCD is a parallelogram and P is a point of bar(CD) . Show that ar(/_\ABP) = ar(/_\ADP+ /_\BCP) .