Question

prove= opposite angles of a cyclic quadrilateral are supplementary

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Answer 1

★Reference of image is in diagram

AnsWer :

$\large\textrm{\underline{Given-}}$

$\bullet$ ABCD is a cyclic quadrilateral. O is the centre.

$\large\textrm{\underline{To \; prove-}}$

$\bullet$ Sum of opposite angles of cyclic quadrilateral is $\bf\ 180^{\circ}$(i.e. <BAD + <BCD = 180°)

$\large\textrm{\underline{Construction:}}$

$\bullet$ Join OB and OD.

$\large\textrm{\underline{ Proof -}}$

$\underline{\dag\:\textsf{According \: to \: given \: in \: question:}}$

$\normalsize\ : \implies\sf\ BAD = \frac{1}{2} \times\ BOD \: \: \:----(eq.1)$

$\:\:\normalsize\sf\ [\therefore\ Angle \: subtended \: by \: arc \: at \: centre \: is \: double \: of \: angle \: on \: circle]$

$\normalsize\ : \implies\sf\ BCD = \frac{1}{2} reflex \: of \: BOD \: \: \: ----(eq.2)$

$\underline{\dag\:\textsf{Adding \: eq.1 \: and \: eq.2, \: we \: get:}}$

$\normalsize\ : \implies\sf\ BAD + BCD = \frac{1}{2}[BOD + reflex \: of \: BOD]$

$\normalsize\ : \implies\sf\ BAD + BCD = \frac{1}{\cancel{2}} \times\ \cancel{360^{\circ}} \\ \\ \normalsize\ : \implies\sf\ BAD + BCD = 180^{\circ}$

$\large\ : \implies{\boxed{\sf{ Hence \: proved!! }}}$