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Prove-that-the-area-of-any-quadrilateral-ABCD-is-s-a-s-b-s-c-s-d-abcd-cos-2-A-C-2-

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Question Number 128178 by bemath last updated on 05/Jan/21
Prove that the area of any quadrilateral ABCD is (√((s−a)(s−b)(s−c)(s−d)−abcd cos^2 (((A+C)/2)))) .
ProvethattheareaofanyquadrilateralABCDis(s−a)(s−b)(s−c)(s−d)−abcdcos2(A+C2).
Answered by mr W last updated on 05/Jan/21
Commented by mr W last updated on 05/Jan/21
let area of ABCD=F s=((a+b+c+d)/2) BD^2 =a^2 +d^2 −2ad cos A=b^2 +c^2 −2bc cos C ad cos A−bc cos C=(1/2)(a^2 +d^2 −b^2 −c^2 ) (ad)^2 cos^2 A+(bc)^2 cos^2 C−2(abcd)cos A cos C=(1/4)(a^2 +d^2 −b^2 −c^2 )^2 ⇒(ad)^2 cos^2 A+(bc)^2 cos^2 C=(1/4)(a^2 +d^2 −b^2 −c^2 )^2 +2(abcd)cos A cos C F=(1/2)(ad sin A+bc sin C) 4F^2 =(ad)^2 sin^2 A+(bc)^2 sin^2 C+2(abcd)sin A sin C 4F^2 =(ad)^2 +(bc)^2 −(ad)^2 cos^2 A−(bc)^2 cos^2 C+2(abcd)sin A sin C 4F^2 =(ad)^2 +(bc)^2 −(1/4)(a^2 +d^2 −b^2 −c^2 )^2 −2(abcd)cos A cos C+2(abcd)sin A sin C 4F^2 =(ad)^2 +(bc)^2 −(1/4)(a^2 +d^2 −b^2 −c^2 )^2 −2(abcd)cos A cos C+2(abcd)sin A sin C 4F^2 =(ad)^2 +(bc)^2 −(1/4)(a^2 +d^2 −b^2 −c^2 )^2 −2(abcd)cos (A+C) 4F^2 =(ad)^2 +(bc)^2 −(1/4)(a^2 +d^2 −b^2 −c^2 )^2 −2(abcd)(2 cos^2 ((A+C)/2)−1) 4F^2 =(ad)^2 +(bc)^2 −(1/4)(a^2 +d^2 −b^2 −c^2 )^2 +2(abcd)−4(abcd) cos^2 ((A+C)/2) 4F^2 =(ad+bc)^2 −(1/4)(a^2 +d^2 −b^2 −c^2 )^2 −4(abcd) cos^2 ((A+C)/2) 16F^2 =(2ad+2bc)^2 −(a^2 +d^2 −b^2 −c^2 )^2 −16(abcd) cos^2 ((A+C)/2) 16F^2 =(2ad+2bc+a^2 +d^2 −b^2 −c^2 )(2ad+2bc−a^2 −d^2 +b^2 +c^2 )−16(abcd) cos^2 ((A+C)/2) 16F^2 =[(a+d)^2 −(b−c)^2 ][(b+c)^2 −(a−d)^2 ]−16(abcd) cos^2 ((A+C)/2) 16F^2 =(a+d+b−c)(a+d−b+c)(b+c+a−d)(b+c−a+d)−16(abcd) cos^2 ((A+C)/2) 16F^2 =(a+d+b+c−2c)(a+d+c+b−2b)(b+c+a+d−2d)(b+c+a+d−2a)−16(abcd) cos^2 ((A+C)/2) 16F^2 =(2s−2c)(2s−2b)(2s−2d)(2s−2a)−16(abcd) cos^2 ((A+C)/2) 16F^2 =16(s−c)(s−b)(s−d)(s−a)−16(abcd) cos^2 ((A+C)/2) F^2 =(s−a)(s−b)(s−c)(s−d)−(abcd) cos^2 ((A+C)/2) ⇒F=(√((s−a)(s−b)(s−c)(s−d)−abcd cos^2 ((A+C)/2)))
letareaofABCD=Fs=a+b+c+d2BD2=a2+d2−2adcosA=b2+c2−2bccosCadcosA−bccosC=12(a2+d2−b2−c2)(ad)2cos2A+(bc)2cos2C−2(abcd)cosAcosC=14(a2+d2−b2−c2)2⇒(ad)2cos2A+(bc)2cos2C=14(a2+d2−b2−c2)2+2(abcd)cosAcosCF=12(adsinA+bcsinC)4F2=(ad)2sin2A+(bc)2sin2C+2(abcd)sinAsinC4F2=(ad)2+(bc)2−(ad)2cos2A−(bc)2cos2C+2(abcd)sinAsinC4F2=(ad)2+(bc)2−14(a2+d2−b2−c2)2−2(abcd)cosAcosC+2(abcd)sinAsinC4F2=(ad)2+(bc)2−14(a2+d2−b2−c2)2−2(abcd)cosAcosC+2(abcd)sinAsinC4F2=(ad)2+(bc)2−14(a2+d2−b2−c2)2−2(abcd)cos(A+C)4F2=(ad)2+(bc)2−14(a2+d2−b2−c2)2−2(abcd)(2cos2A+C2−1)4F2=(ad)2+(bc)2−14(a2+d2−b2−c2)2+2(abcd)−4(abcd)cos2A+C24F2=(ad+bc)2−14(a2+d2−b2−c2)2−4(abcd)cos2A+C216F2=(2ad+2bc)2−(a2+d2−b2−c2)2−16(abcd)cos2A+C216F2=(2ad+2bc+a2+d2−b2−c2)(2ad+2bc−a2−d2+b2+c2)−16(abcd)cos2A+C216F2=[(a+d)2−(b−c)2][(b+c)2−(a−d)2]−16(abcd)cos2A+C216F2=(a+d+b−c)(a+d−b+c)(b+c+a−d)(b+c−a+d)−16(abcd)cos2A+C216F2=(a+d+b+c−2c)(a+d+c+b−2b)(b+c+a+d−2d)(b+c+a+d−2a)−16(abcd)cos2A+C216F2=(2s−2c)(2s−2b)(2s−2d)(2s−2a)−16(abcd)cos2A+C216F2=16(s−c)(s−b)(s−d)(s−a)−16(abcd)cos2A+C2F2=(s−a)(s−b)(s−c)(s−d)−(abcd)cos2A+C2⇒F=(s−a)(s−b)(s−c)(s−d)−abcdcos2A+C2

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