ABCD is a parallelogram. The diagonals AC and BD intersect each other at ‘O’. Prove that `ar(DeltaAOD) = ar(DeltaBOC)` . (Hint: Congruent figures have equal area)

ABCD is trapezium with AB || DC. The diagonal AC and BD intersect at E . If Delta AED ~ Delta BEC . Prove that AD = BC .

ABCD is a rectangle whose diagonals AC and BD intersect at O. If angleOAB= 46^(@) , find angleOBC .

The diagonals of a quadrilateral ABCD intersect each other at point ‘O’ such that (AO)/(BO) = (CO)/(DO) . Prove that ABCD is a trapezium

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 seg AC and seg BD intersects each other at point P and (AP)/(CP)=(BP)/(DP) . Then Prove that DeltaABP~DeltaCDP .

Attempt any Two of the following: In squareABCD ,seg AB ||seg CD . Diagonal AC and BD intersect each other at point P . Prove : (A(DeltaABP))/(A(DeltaCPD))=(AB^(2))/(CD^(2))

ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are . . . . . . .

In a quadrilateral ABCD, it is given that AB |\|CD and the diagonals AC and BD are perpendicular to each other. Show that AD.BC >= AB. CD .