If ABCD is a cyclic quadrilateral, then prove that `AC.BD=AB.CD+BC.AD`

ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P, show that APB = 60^(@) .

In the following figure , ABCD is a cyclic quadrilateral in which AD is parallel to BC . If the bisector of angle A meets BC at point E and given circle at point F, prove that : (i) EF =FC (ii) BF = DF

ABCD is a cyclic quadrilateral , Sides AB and DC produced meet at point E , whereas sides BC and AD produced meet at point F. If angle DCF : angle F : angle E =3 : 5 : 4 find the angles of the cyclic quadrilateral ABCD.

If the sides of a quadrilateral ABCD touch a circle prove that AB+CD=BC+AD.

If ABCD is a quadrilateral such that AB = AD and CB = CD , then prove that AC is the perpendicular bisector of BD .

If ABCD is a quadrilateral, then vec(BA) + vec(BC)+vec(CD) + vec(DA)=

ABCD is a cyclic quadrilateral in which BC is paralleld to AD, angle ADC=110^(@) and angle BAC=50^(@) . Find angle DAC and angle DCA.

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC. or A circle touches all the four sides of a quadrilateral ABCD. Prove that AB+CD=BC+DA.

In Figure, a quadrilateral ABCD is drawn to circumscribe a circle . Prove that AB+CD=BC+AD

In Figure, a quadrilateral ABCD is drawn to circumscribe a circle . Prove that AB+CD=BC+AD