Лице на трапец

[Aneliya]

Диагоналите на трапеца ABCD се пресичат в точка О. Да се докаже, че [tex]S_{ABCD} = (\sqrt{S_{1}} + \sqrt{S_{2}})^{2}[/tex], където S1 е лицето на AOB, а S2 на COD.

[Consigliere-]

Imash trapec ABCD .Neka [tex]S\Delta AOB = S_{1}[/tex] , [tex]S\Delta COD = S_{2}[/tex] , [tex]S\Delta AOD = S_{3}[/tex] , [tex]S\Delta BOC = S_{4}[/tex]
Shte dokajem,che [tex]S_{1}.S_{2}=S{3}.S{4}[/tex]
[tex]S_{1} = \frac{1}{ 2} .OA.OB.sin \varphi[/tex]
[tex]S_{2}= \frac{1}{2 } .OC.OD.sin \varphi[/tex]
Umnojavash i poluchavash
[tex]S_{1}.S_{2} = \frac{1}{4 } .OA.OB.OC.OD.sin^{2}\varphi[/tex] (2)

Po analogichen nachin
[tex]S_{3}.S_{4} = \frac{1}{ 4} .OA.OB.OC.OD.sin^{2}\varphi[/tex] (1)
Ot (1) i (2) [tex]=>[/tex] [tex]S_{1}.S_{2}=S_{3}.S_{4}[/tex] Tova e za proizvolen chetiriugulnik.
Sega shte dokaja che [tex]S_{3}=S_{4}[/tex]
[tex]S_{ABD}=\frac{1}{2 } .AB.DD_{1}=\frac{1}{2 } .AB.h_{trapec}[/tex]
[tex]S_{ABC}=\frac{1}{2 } .AB.CC_{1}=\frac{1}{2 } .AB.h_{trapec}[/tex]
[tex]=> S_{ABD}=S_{ABC}[/tex]
[tex]S_{1}+S_{3}=S_{1}+S_{4}[/tex]
[tex]=> S_{3}=S_{4}[/tex]
ot [tex]S_{1}.S_{2}=S_{3}.S_{4} =>[/tex]
[tex](S_{3})^{2} = S_{1}.S_{2} =>[/tex]
[tex]S_{3} = \sqrt{S_{1}.S_{2}}[/tex]
I sega veche
[tex]S_{ABCD} = S_{1} + S_{2} + 2S_{3}[/tex] , [tex]S_{3} = \sqrt{S_{1}.S_{2}}[/tex]
[tex]S_{ABCD} = S_{1} + 2\sqrt{S_{1}.S_{2}} + S_{2}[/tex]
[tex]S_{ABCD} = (\sqrt{S_{1}} + \sqrt{S_{2}})^{2}[/tex]

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[ganka simeonova]

Consigliere-,много дълго решение.
От подобието на [tex]\Delta COD; \Delta AOB=>\frac{OD}{ OB} =\sqrt{\frac{S_2}{ S_1} }[/tex]
[tex]\frac{S_{AOD}}{ S_{AOB}} =\frac{OD}{ OB} =>S_{AOD}=\sqrt{S_1S_2} =>S_{COB}=\sqrt{S_1S_2}[/tex]
[tex]S_{ABCD}=S_1+2\sqrt{S_1S_2} +S_2=(\sqrt{S_1} +\sqrt{S_2} )^2[/tex]