In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that:- (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE)

ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that: ar(ACB) = ar(ACF)

ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that: ar(AEDF) = ar(ABCDE)

In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle. Also, AL bot DC . Prove that (i) ar (ABCD) = ar (EFCD) (ii) ar (ABCD) = DC × AL

ABCD is a parallelogram and a line through A meets DC at P and BC (produced) at Q. Prove that ar(triangleBPC) = ar(triangleDPQ)

In the 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(APQ)

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

Diagonals AC and BD of quadrilateral ABCD intersect each other at P. Show that ar (APB) xx ar (CPD) = ar (APD)xx ar(BPC) .

The sides AB of parallelogram ABCD produced to any point P. A line through A and parallel to CP meets CB produced at Q and the parallelogram PBQR is completed Show that ar(ABCD) = ar(PBQR)

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

In the fig. ABCD is paralllelogram and E is the mid point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L. prove that AF=2DC