In the figure, ABCD is a quadrilateral. AC is the diagonal and DE || AC and also DE meets BC produced at E. Show that ar(ABCD) = ar `(DeltaABE).`

In the figure, 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)

In, 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)

ABCD is a quadrilateral. Is AB+BC+CD+DA>AC+BD?

In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB .

ABCD is a parallelogram, with AC, BD as diagonals. Then vec AC - vec BD =

In the figure 11.23, E is any point on median AD of a Delta ABC. Show that ar (ABE) = ar (ACE).

In the figure given below, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O , show that (ar(Delta ABC))/(ar(Delta DBC)) = (AO)/(DO) .

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

In the given figure, diagonals AC and BD of quadrilateral ABCD interset at O such that OB = OD. If AB = CD, then show that : (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram .

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