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Question Number 128178 by bemath last updated on 05/Jan/21

$$\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{any}$$$$\mathrm{quadrilateral}\mathrm{ABCD}\mathrm{is}$$$$\sqrt{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})(\mathrm{s}-\mathrm{d})-\mathrm{abcd}\cos {}^2\left(\frac{\mathrm{A}+\mathrm{C}}{2}\right)}.$$

Answered by mr W last updated on 05/Jan/21

Commented by mr W last updated on 05/Jan/21

$$letareaofABCD=F$$$$s=\frac{a+b+c+d}{2}$$$$$$$${BD}^2=a^2+d^2-2ad\cos A=b^2+c^2-2bc\cos C$$$$ad\cos A-bc\cos C=\frac{1}{2}(a^2+d^2-b^2-c^2)$$$${\left(ad\right)}^2{\cos }^{2}A+{\left(bc\right)}^2{\cos }^{2}C-2\left(abcd\right)\cos A\cos C=\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2$$$$\Rightarrow {\left(ad\right)}^2{\cos }^{2}A+{\left(bc\right)}^2{\cos }^{2}C=\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2+2\left(abcd\right)\cos A\cos C$$$$$$$$F=\frac{1}{2}(ad\sin A+bc\sin C)$$$$4F^2={\left(ad\right)}^2{\sin }^{2}A+{\left(bc\right)}^2{\sin }^{2}C+2\left(abcd\right)\sin A\sin C$$$$4F^2={\left(ad\right)}^2+{\left(bc\right)}^2-{\left(ad\right)}^2{\cos }^{2}A-{\left(bc\right)}^2{\cos }^{2}C+2\left(abcd\right)\sin A\sin C$$$$4F^2={\left(ad\right)}^2+{\left(bc\right)}^2-\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2-2\left(abcd\right)\cos A\cos C+2\left(abcd\right)\sin A\sin C$$$$4F^2={\left(ad\right)}^2+{\left(bc\right)}^2-\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2-2\left(abcd\right)\cos A\cos C+2\left(abcd\right)\sin A\sin C$$$$4F^2={\left(ad\right)}^2+{\left(bc\right)}^2-\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2-2\left(abcd\right)\cos (A+C)$$$$4F^2={\left(ad\right)}^2+{\left(bc\right)}^2-\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2-2\left(abcd\right)(2{\cos }^2\frac{A+C}{2}-1)$$$$4F^2={\left(ad\right)}^2+{\left(bc\right)}^2-\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2+2\left(abcd\right)-4\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$4F^2={(ad+bc)}^2-\frac{1}{4}{(a^2+d^2-b^2-c^2)}^2-4\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2={(2ad+2bc)}^2-{(a^2+d^2-b^2-c^2)}^2-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2=(2ad+2bc+a^2+d^2-b^2-c^2)(2ad+2bc-a^2-d^2+b^2+c^2)-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2=[{(a+d)}^2-{(b-c)}^2][{(b+c)}^2-{(a-d)}^2]-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2=(a+d+b-c)(a+d-b+c)(b+c+a-d)(b+c-a+d)-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2=(a+d+b+c-2c)(a+d+c+b-2b)(b+c+a+d-2d)(b+c+a+d-2a)-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2=(2s-2c)(2s-2b)(2s-2d)(2s-2a)-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$16F^2=16(s-c)(s-b)(s-d)(s-a)-16\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$F^2=(s-a)(s-b)(s-c)(s-d)-\left(abcd\right){\cos }^2\frac{A+C}{2}$$$$\Rightarrow F=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd{\cos }^2\frac{A+C}{2}}$$

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