A quadrilateral ABCD is drawn to circumscribe a circle(seeFig).Prove that `AB+CD=AD+BC`

In quadrilateral ABCD AC = AD and AB bisects angleA (see fig. ) DeltaABC ~= DeltaABD . What is the relation between BC and BD ?

In quadrilateral ABCD AC = AD and AB bisects angleA (see fig. ) DeltaABC ~= DeltaABD . What is the relation between BC and BD ?

ABCD is a quadrilateral whose diagonals AC and BD intersect at O, prove that: (AB+BC+CD+DA)>(AC+BD)

In the fig. prove that: AD+AB+BC+CD>2AC.

ABCD is a quadrilateral whose diagonals AC and BD intersect at O, prove that: (AB+BC+CD+DA)>2(AC+BD)

The area of a cyclic quadrilateral ABCD is (3sqrt3)/4 . The radius of the circle circumscribing cyclic quadrilateral is 1.If AB =1 and BD =sqrt3 , then BC*CD is equal to

If bisectors of opposiste angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

In the fig. ABCD is a quadrilateral in which AB is the largest and CD is the shortest side. Prove that: angleC>angleA

In the fig. ABCD is a quadrilateral in which AB is the largest and CD is the shortest side. Prove that: angleD>angleB