In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

ar(`Delta`AEDF) = ar(`Delta`ABCDE)

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P . Prove that ar(triangleABP)=ar("qaud. ABCD") .

ABCD is a trapezium with AB||DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar( Delta ADX) = ar( Delta ACY).

In the figure, ABCD is a quadrilateral. A line DP drawn parallel to diagonal AC from point D, meet BC produced at P. Prove that: are of Delta ABP = " area of " square ABCD

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (DeltaAPQ) .

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar(DeltaADF) = ar (ABFC) .

ABCD is a trapezium in which AB I I DC. DC is produced to E such tha t CE= AB, prove that ar( Delta ABD) = ar( Delta BCE).

The side AB of a parallelogram ABCD is produced to any point E. A line through A and parallel to CE meets CB produced at G and then parallelogram EBGF is completed (see the figure). Prove that area of ||gm ABCD = " area of "||gm BEFG .

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed as shown in Figure. Show that ar(||^(gm) ABCD)=ar (||^(gm) PBQR)

ABCD is a quadrilateral. The straight line through C parallel to the diagonal DB intersects AB produced at E. Prove that the ar(quad. ABCD) = ar (triangleADE) .

P is the mid-point of side AB of a parallelogram ABCD .A Line through B parallel to PD meets DC at Q and AD produced at R. Prove that ( i )AR=2BC (ii) BR=2BQ.