Assume A , B , C , D A,B,C,D A , B , C , D p , q , r , s p,q,r,s p , q , r , s A D , A B , B C , C D , AD,AB,BC,CD, A D , A B , BC , C D ,
( Area of A B C D ) ≡ K = ( Area of △ A D B ) + ( Area of △ B D C ) = p q sin A 2 + r s sin C 2 . \begin{aligned}
(\text{Area of } ABCD) \equiv K
&=(\text{Area of } \triangle ADB)+(\text{Area of } \triangle BDC)\\\\
&=\frac { pq \sin A }{ 2 } +\frac { rs \sin C }{ 2 }.
\end{aligned} ( Area of A BC D ) ≡ K = ( Area of △ A D B ) + ( Area of △ B D C ) = 2 pq sin A + 2 rs sin C .
Since the quadrilateral is cyclic, sin A = sin C , \sin A=\sin C, sin A = sin C ,
K = p q sin A 2 + r s sin A 2 K 2 = 1 4 ( p q + r s ) 2 sin 2 A 4 K 2 = ( p q + r s ) 2 ( 1 − cos 2 A ) = ( p q + r s ) 2 − ( p q + r s ) 2 ( cos 2 A ) . ( 1 ) \begin{aligned}
K&=\frac { pq\sin A }{ 2 } +\frac { rs\sin A }{ 2 } \\
K^2&=\frac { 1 }{ 4 } { ( pq+rs ) }^{ 2 }\sin ^{ 2 }{ A } \\
4K^{ 2 }&={ \left( pq+rs \right) }^{ 2 }\big( 1-\cos ^{ 2 }{ A } \big) \\
&={ \left( pq+rs \right) }^{ 2 }-{ \left( pq+rs \right) }^{ 2 }\big( \cos ^{ 2 }{ A } \big). \qquad (1)
\end{aligned} K K 2 4 K 2 = 2 pq sin A + 2 rs sin A = 4 1 ( pq + rs ) 2 sin 2 A = ( pq + rs ) 2 ( 1 − cos 2 A ) = ( pq + rs ) 2 − ( pq + rs ) 2 ( cos 2 A ) . ( 1 )
Solving for common side D B DB D B A D B ADB A D B B D C BDC B D C
p 2 + q 2 − 2 p q cos A = r 2 + s 2 − 2 r s cos C . { p }^{ 2 }+{ q }^{ 2 }-2pq\cos A={ r }^{ 2 }+{ s }^{ 2 }-2rs\cos C. p 2 + q 2 − 2 pq cos A = r 2 + s 2 − 2 rs cos C .
As angles A A A C C C cos A = − cos C \cos A=-\cos C cos A = − cos C
2 ( p q + r s ) cos A = p 2 + q 2 − r 2 − s 2 . 2(pq+rs)\cos A={ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }. 2 ( pq + rs ) cos A = p 2 + q 2 − r 2 − s 2 .
Substituting this into the equation of area ( 1 ) , (1), ( 1 ) ,
16 K 2 = 4 ( p q + r s ) 2 − ( p 2 + q 2 − r 2 − s 2 ) 2 . 16{ K }^{ 2 }={ 4(pq+rs) }^{ 2 }-{ \big({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\big) }^{ 2 }. 16 K 2 = 4 ( pq + rs ) 2 − ( p 2 + q 2 − r 2 − s 2 ) 2 .
The RHS of the equation is of the form a 2 − b 2 { a }^{ 2 }-{ b }^{ 2 } a 2 − b 2
[ 2 ( p q + r s ) − ( p 2 + q 2 − r 2 − s 2 ) ] [ 2 ( p q + r s ) + ( p 2 + q 2 − r 2 − s 2 ) ] = [ ( r + s ) 2 − ( p − q ) 2 ] [ ( p + q ) 2 − ( r − s ) 2 ] = ( p + q + r − s ) ( p + q − r + s ) ( p − q + r + s ) ( − p + q + r + s ) . \begin{aligned}
\Big[2(pq+rs)-\big({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\big)\Big]\Big[2(pq+rs)+\big({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\big)\Big]
&=\big[ { \left( r+s \right) }^{ 2 }-{ \left( p-q \right) }^{ 2 } \big] \big[ { \left( p+q \right) }^{ 2 }-{ \left( r-s \right) }^{ 2 } \big] \\ \\
&=(p+q+r-s)(p+q-r+s)(p-q+r+s)(-p+q+r+s).
\end{aligned} [ 2 ( pq + rs ) − ( p 2 + q 2 − r 2 − s 2 ) ] [ 2 ( pq + rs ) + ( p 2 + q 2 − r 2 − s 2 ) ] = [ ( r + s ) 2 − ( p − q ) 2 ] [ ( p + q ) 2 − ( r − s ) 2 ] = ( p + q + r − s ) ( p + q − r + s ) ( p − q + r + s ) ( − p + q + r + s ) .
So, the area of the cyclic quadrilateral with sides a , b , c , d a,b,c,d a , b , c , d
K = 1 4 ( a + b + c − d ) ( a + b − c + d ) ( a − b + c + d ) ( − a + b + c + d ) . □ K=\frac { 1 }{ 4 } \sqrt { (a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) }.\ _\square K = 4 1 ( a + b + c − d ) ( a + b − c + d ) ( a − b + c + d ) ( − a + b + c + d ) . □
Example of a Cyclic Quadrilateral