Četverougao Bretschneider's-ova formula se koristi u geometriji za određivanje površine četvorougla , i glasi
P = ( s − a ) ( s − b ) ( s − c ) ( s − d ) − a b c d cos 2 α + γ 2 , {\displaystyle P={\sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}}},} pri čemu su, a , b , c i d stranice četvorougla, s je polovina obima četvorougla, a α {\displaystyle \alpha \,} γ {\displaystyle \gamma \,}
Bretschneider's-ova formula daje površinu četvorougla bez obzira da li je on tetivan ili nije.
Dokaz Ako se površina četvorougla označi sa P , onda važi
P = P △ B D C + P △ A D B = 1 2 a b sin γ + 1 2 c d sin α {\displaystyle {\begin{aligned}P=P_{\triangle BDC}+P_{\triangle ADB}={\tfrac {1}{2}}ab\sin \gamma +{\tfrac {1}{2}}cd\sin \alpha \end{aligned}}} 4 P 2 = ( a b ) 2 sin 2 γ + ( c d ) 2 sin 2 α + 2 a b c d sin α sin γ . {\displaystyle 4P^{2}=(ab)^{2}\sin ^{2}\gamma +(cd)^{2}\sin ^{2}\alpha +2abcd\sin \alpha \sin \gamma .\,} Prema kosinusnoj teoremi
a 2 + b 2 − 2 a b cos γ = c 2 + d 2 − 2 c d cos α , {\displaystyle a^{2}+b^{2}-2ab\cos \gamma =c^{2}+d^{2}-2cd\cos \alpha ,\,} Pošto su obe strane jednake kvadratu dužine dijagonale BD i 1 4 ( c 2 + d 2 − a 2 − b 2 ) 2 = ( a b ) 2 cos 2 γ + ( c d ) 2 cos 2 α − 2 a b c d cos α cos γ . {\displaystyle {\tfrac {1}{4}}(c^{2}+d^{2}-a^{2}-b^{2})^{2}=(ab)^{2}\cos ^{2}\gamma +(cd)^{2}\cos ^{2}\alpha -2abcd\cos \alpha \cos \gamma .\,} 4 P 2 + 1 4 ( c 2 + d 2 − a 2 − b 2 ) 2 = ( a b ) 2 + ( c d ) 2 − 2 a b c d cos ( α + γ ) . {\displaystyle 4P^{2}+{\tfrac {1}{4}}(c^{2}+d^{2}-a^{2}-b^{2})^{2}=(ab)^{2}+(cd)^{2}-2abcd\cos(\alpha +\gamma ).\,} 16 P 2 = 4 ( a 2 b 2 + c 2 d 2 ) − ( c 2 + d 2 − a 2 − b 2 ) 2 − 8 a b c d cos ( α + γ ) . {\displaystyle 16P^{2}=4(a^{2}b^{2}+c^{2}d^{2})-(c^{2}+d^{2}-a^{2}-b^{2})^{2}-8abcd\cos(\alpha +\gamma ).\,} 16 P 2 = 4 ( a b + c d ) 2 − ( c 2 + d 2 − a 2 − b 2 ) 2 − 8 a b c d [ 1 + cos ( α + γ ) ] . {\displaystyle 16P^{2}=4(ab+cd)^{2}-(c^{2}+d^{2}-a^{2}-b^{2})^{2}-8abcd[1+\cos(\alpha +\gamma )].\,} 16 P 2 = [ 2 ( a b + c d ) − ( c 2 + d 2 − a 2 − b 2 ) ] [ 2 ( a b + c d ) + ( c 2 + d 2 − a 2 − b 2 ) ] − 16 a b c d cos 2 α + γ 2 , {\displaystyle 16P^{2}=[2(ab+cd)-(c^{2}+d^{2}-a^{2}-b^{2})][2(ab+cd)+(c^{2}+d^{2}-a^{2}-b^{2})]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}},\,} 16 P 2 = [ ( a + b ) 2 − ( c − d ) 2 ] [ ( c + d ) 2 − ( a − b ) 2 ] − 16 a b c d cos 2 α + γ 2 . {\displaystyle 16P^{2}=[(a+b)^{2}-(c-d)^{2}][(c+d)^{2}-(a-b)^{2}]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}.\,} 16 P 2 = ( a + b + c − d ) ( a + b − c + d ) ( c + d + a − b ) ( c + d − a + b ) − 16 a b c d cos 2 α + γ 2 . {\displaystyle 16P^{2}=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}.} Uzevši u obzir za poluobim četverougla
s = a + b + c + d 2 , {\displaystyle s={\frac {a+b+c+d}{2}},} imamo
16 P 2 = 16 ( s − a ) ( s − b ) ( s − c ) ( s − d ) − 16 a b c d cos 2 α + γ 2 {\displaystyle 16P^{2}=16(s-a)(s-b)(s-c)(s-d)-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}} Bretschneider's-ova formulaje uopštenje formule Bramagupte za površinu tetivnog četvorougla, a ova je uopštenje Heronovog obrasca koji se koristi za izračunavanje površine trougla.
Izvori Weisstein, Eric W: Bretschneider's Formula MathWorld Bretschneider's Formula 
![{\displaystyle 16P^{2}=4(ab+cd)^{2}-(c^{2}+d^{2}-a^{2}-b^{2})^{2}-8abcd[1+\cos(\alpha +\gamma )].\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c8522a330ad9e3791d58d10ff32485453eaf9d9)
![{\displaystyle 16P^{2}=[2(ab+cd)-(c^{2}+d^{2}-a^{2}-b^{2})][2(ab+cd)+(c^{2}+d^{2}-a^{2}-b^{2})]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}},\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c6f1d7bd146e9bee1c1e5b0df9a7ed7e67a503d)
![{\displaystyle 16P^{2}=[(a+b)^{2}-(c-d)^{2}][(c+d)^{2}-(a-b)^{2}]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/049dd807f0ac5ef9be248dc9466c440bcbd2c004)
