A,B,C,D are the angles of a cyclic quadrilateral prove that sinA+sinB=sinC+sind
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A+C=180°
A=180°- C
sinA=sin(180-C)
sinA=sinC ......(1)
B+D=180°
B=180-D
sinB=sin(180-D)
sinB= sinD ........(2)
A=180°- C
sinA=sin(180-C)
sinA=sinC ......(1)
B+D=180°
B=180-D
sinB=sin(180-D)
sinB= sinD ........(2)
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