In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If DBC55 and BAC45, find BCD.

Therefore, DAB = CAD + BAC

= 55 + 45 = 100

But DAB + BCD = 180 (Opposite angles of a cyclic quadrilateral)

So, BCD = 180 100 = 80

LBAC = LBDC.........(angles subtended bu the same chord)

Hence, LBDC = 45^O

NOW, in triangle BDC,

LDBC + LBDC + LBCD = 180^{0............}(Angle sum prop.)

= 55⁰ + 45⁰ + LBCD = 180⁰

= LBCD = 180 - (55 + 45)

= LBCD = 80⁰

hope this helps you............