Question Number 128178 by bemath last updated on 05/Jan/21
$$\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{any}\: \\ $$$$\mathrm{quadrilateral}\:\mathrm{ABCD}\:\mathrm{is}\: \\ $$$$\:\sqrt{\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)\left(\mathrm{s}−\mathrm{d}\right)−\mathrm{abcd}\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{A}+\mathrm{C}}{\mathrm{2}}\right)}\:. \\ $$
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)))](https://www.tinkutara.com/question/Q128195.png)
$${let}\:{area}\:{of}\:{ABCD}={F} \\ $$$${s}=\frac{{a}+{b}+{c}+{d}}{\mathrm{2}} \\ $$$$ \\ $$$${BD}^{\mathrm{2}} ={a}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{ad}\:\mathrm{cos}\:{A}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:{C} \\ $$$${ad}\:\mathrm{cos}\:{A}−{bc}\:\mathrm{cos}\:{C}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$$\left({ad}\right)^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:{A}+\left({bc}\right)^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:{C}−\mathrm{2}\left({abcd}\right)\mathrm{cos}\:{A}\:\mathrm{cos}\:{C}=\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({ad}\right)^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:{A}+\left({bc}\right)^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:{C}=\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\left({abcd}\right)\mathrm{cos}\:{A}\:\mathrm{cos}\:{C} \\ $$$$ \\ $$$${F}=\frac{\mathrm{1}}{\mathrm{2}}\left({ad}\:\mathrm{sin}\:{A}+{bc}\:\mathrm{sin}\:{C}\right) \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{A}+\left({bc}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{C}+\mathrm{2}\left({abcd}\right)\mathrm{sin}\:{A}\:\mathrm{sin}\:{C} \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} −\left({ad}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{A}−\left({bc}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{C}+\mathrm{2}\left({abcd}\right)\mathrm{sin}\:{A}\:\mathrm{sin}\:{C} \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({abcd}\right)\mathrm{cos}\:{A}\:\mathrm{cos}\:{C}+\mathrm{2}\left({abcd}\right)\mathrm{sin}\:{A}\:\mathrm{sin}\:{C} \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({abcd}\right)\mathrm{cos}\:{A}\:\mathrm{cos}\:{C}+\mathrm{2}\left({abcd}\right)\mathrm{sin}\:{A}\:\mathrm{sin}\:{C} \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({abcd}\right)\mathrm{cos}\:\left({A}+{C}\right) \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({abcd}\right)\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\left({abcd}\right)−\mathrm{4}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{4}{F}^{\mathrm{2}} =\left({ad}+{bc}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\left(\mathrm{2}{ad}+\mathrm{2}{bc}\right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\left(\mathrm{2}{ad}+\mathrm{2}{bc}+{a}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left(\mathrm{2}{ad}+\mathrm{2}{bc}−{a}^{\mathrm{2}} −{d}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\left[\left({a}+{d}\right)^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \right]\left[\left({b}+{c}\right)^{\mathrm{2}} −\left({a}−{d}\right)^{\mathrm{2}} \right]−\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\left({a}+{d}+{b}−{c}\right)\left({a}+{d}−{b}+{c}\right)\left({b}+{c}+{a}−{d}\right)\left({b}+{c}−{a}+{d}\right)−\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\left({a}+{d}+{b}+{c}−\mathrm{2}{c}\right)\left({a}+{d}+{c}+{b}−\mathrm{2}{b}\right)\left({b}+{c}+{a}+{d}−\mathrm{2}{d}\right)\left({b}+{c}+{a}+{d}−\mathrm{2}{a}\right)−\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\left(\mathrm{2}{s}−\mathrm{2}{c}\right)\left(\mathrm{2}{s}−\mathrm{2}{b}\right)\left(\mathrm{2}{s}−\mathrm{2}{d}\right)\left(\mathrm{2}{s}−\mathrm{2}{a}\right)−\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\mathrm{16}{F}^{\mathrm{2}} =\mathrm{16}\left({s}−{c}\right)\left({s}−{b}\right)\left({s}−{d}\right)\left({s}−{a}\right)−\mathrm{16}\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$${F}^{\mathrm{2}} =\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)\left({s}−{d}\right)−\left({abcd}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}} \\ $$$$\Rightarrow{F}=\sqrt{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)\left({s}−{d}\right)−{abcd}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}+{C}}{\mathrm{2}}} \\ $$