In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

We know that, Area of a triangle is half that of a parallelogram on the same base and between the same parallel lines. ∆APD and || gm ABCD are on the same base AD and between the same || lines AD and BC, ∴ Area of ∆APD = Area of || gm ABCD .......(i) From figure, Area of ||gm ABCD = Area of ∆APD + Area of ∆ABP + Area of ∆DPC Dividing the above equation by 2 we get, From (i), Hence, proved that Area of ∆APD = Area of ∆ABP + Area of ∆DPC.

In figure (1) given below, AD is the median of ∆ABC and P is any point on AD. Prove that
(i) Area of ∆PBD = area of ∆PDC.
(ii) Area of ∆ABP = area of ∆ACP.

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In the figure (2) given below, DE || BC. Prove that
(i) area of ∆ACD = area of ∆ABE
(ii) area of ∆OBD = area of ∆OCE.

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Mathematics

In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

Theorems on Area

Related Questions

In figure (1) given below, AD is the median of ∆ABC and P is any point on AD. Prove that
(i) Area of ∆PBD = area of ∆PDC.
(ii) Area of ∆ABP = area of ∆ACP.

In the figure (2) given below, DE || BC. Prove that
(i) area of ∆ACD = area of ∆ABE
(ii) area of ∆OBD = area of ∆OCE.

Mathematics

In figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that, Area of ∆ABP + area of ∆DPC = Area of ∆APD.

Theorems on Area

Related Questions

In figure (1) given below, AD is the median of ∆ABC and P is any point on AD. Prove that(i) Area of ∆PBD = area of ∆PDC.(ii) Area of ∆ABP = area of ∆ACP.

In the figure (2) given below, DE || BC. Prove that(i) area of ∆ACD = area of ∆ABE(ii) area of ∆OBD = area of ∆OCE.

In figure (1) given below, AD is the median of ∆ABC and P is any point on AD. Prove that
(i) Area of ∆PBD = area of ∆PDC.
(ii) Area of ∆ABP = area of ∆ACP.

In the figure (2) given below, DE || BC. Prove that
(i) area of ∆ACD = area of ∆ABE
(ii) area of ∆OBD = area of ∆OCE.