Question

prove that the sum of a pair of opposite angles of a quadrilateral is 180 degree, the quadrilateral is cyclic​

Answer 1

$\large{\boxed{\red{\boxed{\underline{\green{\mathfrak{Given :-}}}}}}}$

$\bold{\red{\textsf{ABCD is a cyclic quadrilateral, With Centre O.}}}$

$\large{\boxed{\red{\boxed{\underline{\green{\mathfrak{To \: Prove :-}}}}}}}$

$\bold{\red{\textsf{Sum of pair of Opposite angles is 180°}}}$

$\bold{\mathsf{i.e, \: \: • \angle BAD + \angle BCD = 180\degree}}$

$\bold{\mathsf{\: \: \: \: \: \: \: \: • \angle ABC + \angle ADC = 180\degree}}$

$\large{\boxed{\red{\boxed{\underline{\green{\mathfrak{Proof :-}}}}}}}$

$\boxed{\textrm{\pink{Hence, Angles in same segment are equal}}}$

$\textrm{\pink{So,}}$

$\longrightarrow \textsf{\purple{In Chord AB,}}$

$\boxed{\mathsf{\blue{\angle 5 = \angle 8 \: \: ....(1)}}}$

$\longrightarrow \textsf{\purple{In Chord BC,}}$

$\boxed{\mathsf{\blue{\angle 1 = \angle 6 \: \: ....(2)}}}$

$\longrightarrow \textsf{\purple{In Chord CD,}}$

$\boxed{\mathsf{\blue{\angle 2 = \angle 4 \: \: ....(3)}}}$

$\longrightarrow \textsf{\purple{In Chord AD,}}$

$\boxed{\mathsf{\blue{\angle 7 = \angle 3 \: \: ....(4)}}}$

$\textrm{\pink{By Angle Sum Property of Quadrilateral}}$

$\longrightarrow{\boxed{\mathsf{\purple{\angle A + \angle B + \angle C + \angle D = 360\degree}}}}$

$\therefore{\mathsf{\blue{\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 7}}}$

$\sf \blue{+ \angle 8 + \angle 5 + \angle 6 = 360\degree}$

$\implies \mathsf{\blue{(\angle 1 + \angle 2 + \angle 7 + \angle 8) + }}$

$\sf \blue{(\angle 3 + \angle 4 + \angle 5 + \angle 6) = 360\degree}$

$\longrightarrow{\pink{\textrm{From (1), (2), (3) and (4),}}}$$\pink{\textrm{We get,}}$

$\therefore \mathsf{\purple{(\angle 1 + \angle 2 + \angle 7 + \angle 8) +}}$

$\sf \purple{(\angle 7 + \angle 2 + \angle 8 + \angle 1) = 360\degree}$

$\implies \mathsf{\blue{2(\angle 1 + \angle 2 + \angle 7 + \angle 8) = 360\degree}}$

$\implies \sf \blue{\angle 1 + \angle 2 + \angle 7 + \angle 8 = \cancel{\frac{360\degree}{2}}}$

$\implies \sf \blue{\angle 1 + \angle 2 + \angle 7 + \angle 8 = 180\degree}$

$\implies \sf \blue{(\angle 1 + \angle 2) + (\angle 7 + \angle 8) = 180\degree}$

$\pink{\mathsf{Hence, \: \angle 1 + \angle 2 = \angle BAD}}$

$\pink{\mathsf{And, \: \angle 7 + \angle 8 = \angle BCD}}$

$\pink{\textrm{Therefore,}}$

$\boxed{\purple{\mathsf{ \angle BAD + \angle BCD = 180\degree}}}$

$\pink{\textrm{Similarly,}}$

$\boxed{\purple{\mathsf{ \angle ABC + \angle ADC = 180\degree}}}$

$\green{\underline{\texttt{Hence, Proved.}}}$