 Hi and welcome to the session. Let us discuss the following question. Question says, the diagonals of a quadrilateral ABCD intersect each other at 0.4 such that AO upon DO is equal to CO upon DO. Show that ABCDs are trapezius. First of all let us recall basic proportionality theorem. Basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in two distinct points then it divides the other two sides in same ratio. If we have given a triangle ABC and DE is parallel to BC then AD upon DB is equal to AE upon EC. Now let us understand converse of basic proportionality theorem. Converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio then the line is parallel to the third side of a triangle. If we are given a triangle ABC and DE is a line which divides AB and AC in the same ratio then DE is parallel to BC. This is the key idea to solve the given question. Now let us start the solution. First of all let us draw quadrilateral ABCD in which diagonals AC and BD intersect each other at 0 such that AO upon BO is equal to CO upon DO. So we can write given in quadrilateral ABCD AC and BD intersect each other at 0 such that AO upon BO is equal to CO upon DO. Now we have to prove that ABCD is a trapezium. Now before starting the proof we will do some construction. Through O draw a line parallel to AB such that it intersects AD at E. Now this is our required construction. Let us now start the proof. First of all let us see triangle ADB in triangle ADB. OE is parallel to AB by construction. Now by applying basic proportionality theorem we get DE upon EA is equal to DO upon OB. Now taking reciprocal of both the ratios we get EA upon DE is equal to OB upon DO. Now let us name this expression as 1. We know that AO upon BO is equal to CO upon DO. This is given in the question. Now multiplying both sides by BO upon CO we get AO upon CO is equal to BO upon DO. Or we can rewrite this as AO upon CO is equal to OB upon DO. BO can be written as OB. Now let us name this expression as 2. Now OB upon DO is equal to EA upon DE and OB upon DO is equal to AO upon CO. So from 1 and 2 we get EA upon DE is equal to AO upon CO. Since both the ratios are equal to OB upon DO, EA upon DE is equal to AO upon CO implies by a converse of basic proportionality theorem OE is parallel to DC. OE is a line which is dividing AD and AC in the same ratio. So OE is parallel to CD. So we can write in triangle ADC EA upon DE is equal to AO upon CO. This we have proved above. So by converse of basic proportionality theorem we get OE is parallel to CD. Also OE is parallel to AV by construction. Now we know two lines parallel to same given line are parallel to each other. Now CD and AV both are parallel to OE. So this implies CD is parallel to AV. Two lines parallel to same given line are parallel to each other. Now we get CD is parallel to AV. This implies AV-CD is a trapezium. We know quadrilateral in which one pair of opposite sides is parallel is a trapezium. So we can write. Therefore quadrilateral AV-CD is a trapezium. So this is our required answer. This completes the session. Hope you understood the session. Have a nice day.