Prove that opposite sides of quadrilateral …

Given: ABCD is a quadrilateral circumscribing a circle whose centre is O.
To prove:

1. {tex}\angle{/tex}AOB + {tex}\angle{/tex}COD = 180^o
2. {tex}\angle{/tex}BOC + {tex}\angle{/tex}AOD = 180^o

Construction: Join OP, OQ, OR and OS.
 
Proof: Since tangents from an external point to a circle are equal.
{tex}\therefore{/tex} AP = AS,
BP = BQ ........ (i)
CQ = CR
DR = DS
In {tex}\triangle{/tex}OBP and {tex}\triangle{/tex}OBQ,
OP = OQ [Radii of the same circle]
OB = OB [Common]
BP = BQ [From eq. (i)]
{tex}\therefore{/tex} {tex}\triangle{/tex}OPB {tex}\cong{/tex} {tex}\triangle{/tex}OBQ [By SSS congruence criterion]
{tex}\therefore{/tex} {tex}\angle{/tex}1 = {tex}\angle{/tex}2 [By C.P.C.T.]
Similarly, {tex}\angle{/tex}3 = {tex}\angle{/tex}4, {tex}\angle{/tex}5 = {tex}\angle{/tex}6, {tex}\angle{/tex}7 = {tex}\angle{/tex}8
Since, the sum of all the angles round a point is equal to 360^o.
{tex}\therefore{/tex} {tex}\angle{/tex}1 + {tex}\angle{/tex}2 + {tex}\angle{/tex}3 + {tex}\angle{/tex}4+ {tex}\angle{/tex}5 + {tex}\angle{/tex}6 + {tex}\angle{/tex}7 + {tex}\angle{/tex}8 = 360^o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}1 + {tex}\angle{/tex}1 + {tex}\angle{/tex}4 + {tex}\angle{/tex}4+ {tex}\angle{/tex}5 + {tex}\angle{/tex}5 + {tex}\angle{/tex}8 + {tex}\angle{/tex}8 = 360^o
{tex}\Rightarrow{/tex}  2 ({tex}\angle{/tex}1 + {tex}\angle{/tex}4 + {tex}\angle{/tex}5 + {tex}\angle{/tex}8)  = 360^o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}1 + {tex}\angle{/tex}4 + {tex}\angle{/tex}5 + {tex}\angle{/tex}8 = 180^o
{tex}\Rightarrow{/tex} ({tex}\angle{/tex}1 + {tex}\angle{/tex}5)  + ({tex}\angle{/tex}4 + {tex}\angle{/tex}8) = 180^o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}AOB + {tex}\angle{/tex}COD = 180^o
Similarly we can prove that
{tex}\angle{/tex}BOC + {tex}\angle{/tex}AOD = 180^o