Teorema Falesa odna iz teorem planimetriyi U matematichnij literaturi krayin kolishnogo Radyanskogo Soyuzu vidoma yak teorema Falesa ta uzagalnena teorema Falesa teorema pro proporcijni vidrizki U yevropejskij literaturi teoremoyu Falesa najchastishe nazivayut inshu teoremu Zmist 1 Istoriya 2 Formulyuvannya 3 Dovedennya teoremi Falesa 4 Dovedennya uzagalnenoyi teoremi Falesa 5 Literatura 6 PosilannyaIstoriya RedaguvatiTeorema Falesa nalezhit davnogreckomu matematiku i filosofu Falesu Miletskomu Za legendoyu Fales Miletskij znahodiv visotu piramidi Heopsa vimiryuyuchi dovzhinu yiyi tini na zemli ta dovzhinu tini palici vimiryuvanoyi visoti Najpershe pismove dovedennya ciyeyi teoremi podano v knizi Nachala kniga VI boti go u ffFormulyuvannya RedaguvatiTeorema Falesa yaksho paralelni pryami sho peretinayut dvi zadani pryami a i b vidtinayut na odnij pryamij rivni vidrizki to voni vidtinayut rivni vidrizki j na inshij pryamij A 1 A 2 A 2 A 3 displaystyle A 1 A 2 A 2 A 3 nbsp to B 1 B 2 B 2 B 3 displaystyle B 1 B 2 B 2 B 3 nbsp Uzagalnena teorema Falesa paralelni pryami sho peretinayut dvi zadani pryami a i b vidtinayut na nih proporcijni vidrizki A 1 A 2 B 1 B 2 A 2 A 3 B 2 B 3 A 1 A 3 B 1 B 3 displaystyle frac A 1 A 2 B 1 B 2 frac A 2 A 3 B 2 B 3 frac A 1 A 3 B 1 B 3 nbsp Dovedennya teoremi Falesa Redaguvati nbsp Malyunok 1Nehaj dano paralelni pryami A 1 B 1 displaystyle A 1 B 1 parallel nbsp A 2 B 2 displaystyle A 2 B 2 parallel nbsp A 3 B 3 displaystyle A 3 B 3 nbsp yaki peretinayut pryami a displaystyle a nbsp i b displaystyle b nbsp prichomu A 1 A 2 A 2 A 3 displaystyle A 1 A 2 A 2 A 3 nbsp divitis pravoruch Malyunok 1 Cherez tochki A 1 displaystyle A 1 nbsp i A 2 displaystyle A 2 nbsp provedeno pryami A 1 M displaystyle A 1 M nbsp i A 2 K displaystyle A 2 K nbsp paralelni pryamij b displaystyle b nbsp A 1 A 2 M A 2 A 3 K displaystyle vartriangle A 1 A 2 M vartriangle A 2 A 3 K nbsp za drugoyu oznakoyu rivnosti trikutnikiv oskilki 1 A 1 A 2 A 2 A 3 displaystyle A 1 A 2 A 2 A 3 nbsp za umovoyu 2 A 1 A 2 M A 2 A 3 K displaystyle angle A 1 A 2 M angle A 2 A 3 K nbsp vidpovidni kuti pri paralelnih pryamih M A 2 displaystyle MA 2 nbsp i K A 3 displaystyle KA 3 nbsp 3 A 2 A 1 M A 3 A 2 K displaystyle angle A 2 A 1 M angle A 3 A 2 K nbsp vidpovidni kuti pri paralelnih pryamih A 1 M displaystyle A 1 M nbsp i A 2 K displaystyle A 2 K nbsp Z rivnosti trikutnikiv A 1 A 2 M A 2 A 3 K displaystyle vartriangle A 1 A 2 M vartriangle A 2 A 3 K nbsp displaystyle Rightarrow nbsp A 1 M displaystyle A 1 M nbsp A 2 K displaystyle A 2 K nbsp yak vidpovidni storoni rivnih trikutnikiv nbsp Malyunok 2Z pobudovi Malyunok 1 chotirikutnik A 1 B 1 B 2 M displaystyle A 1 B 1 B 2 M nbsp paralelogram tomu A 1 M B 1 B 2 displaystyle A 1 M B 1 B 2 nbsp Z pobudovi Malyunok 1 chotirikutnik A 2 B 2 B 3 K displaystyle A 2 B 2 B 3 K nbsp paralelogram tomu A 2 K B 2 B 3 displaystyle A 2 K B 2 B 3 nbsp Zvidsi A 1 M A 2 K displaystyle A 1 M A 2 K nbsp i B 1 B 2 B 2 B 3 displaystyle B 1 B 2 B 2 B 3 nbsp Dovedennya uzagalnenoyi teoremi Falesa RedaguvatiNehaj pryami a displaystyle a nbsp i b displaystyle b nbsp peretinayut paralelni pryami u tochkah A 1 A 2 A 3 displaystyle A 1 A 2 A 3 nbsp i B 1 B 2 B 3 displaystyle B 1 B 2 B 3 nbsp vidpovidno divitis pravoruch Malyunok 2 Dovedemo sho A 1 A 2 A 1 A 3 B 1 B 2 B 1 B 3 displaystyle frac A 1 A 2 A 1 A 3 frac B 1 B 2 B 1 B 3 nbsp dlya vipadku koli isnuye vidrizok takoyi dovzhini d displaystyle delta nbsp yakij mozhna vidklasti cile chislo raziv na vidrizku A 1 A 3 displaystyle A 1 A 3 nbsp i A 1 A 2 displaystyle A 1 A 2 nbsp Nehaj A 1 A 3 n d displaystyle A 1 A 3 n delta nbsp A 1 A 2 m d displaystyle A 1 A 2 m delta nbsp i n gt m displaystyle n gt m nbsp Podilimo vidrizok A 1 A 3 displaystyle A 1 A 3 nbsp na n displaystyle n nbsp rivnih chastin dovzhinoyu d displaystyle delta nbsp tochka A 2 displaystyle A 2 nbsp odna z tochok podilu Cherez tochki podilu provedemo pryami paralelni A 3 B 3 displaystyle A 3 B 3 nbsp Za teoremoyu Falesa ci pryami dilyat vidrizok B 1 B 3 displaystyle B 1 B 3 nbsp na rivni vidrizki deyakoyi dovzhini d 1 displaystyle delta 1 nbsp Otrimayemo B 1 B 3 n d 1 displaystyle B 1 B 3 n delta 1 nbsp B 1 B 2 m d 1 displaystyle B 1 B 2 m delta 1 nbsp A 1 A 2 A 1 A 3 m d n d m n displaystyle frac A 1 A 2 A 1 A 3 frac m delta n delta frac m n nbsp i B 1 B 2 B 1 B 3 m d 1 n d 1 m n displaystyle frac B 1 B 2 B 1 B 3 frac m delta 1 n delta 1 frac m n nbsp displaystyle Longrightarrow nbsp A 1 A 2 A 1 A 3 B 1 B 2 B 1 B 3 displaystyle frac A 1 A 2 A 1 A 3 frac B 1 B 2 B 1 B 3 nbsp Literatura RedaguvatiPogoryelov O V Geometriya Planimetriya Pidruch dlya 7 9 kl zagalnoosvit navch zakl 7 me vid K Shkolyar 2004 S 85 87 Posilannya RedaguvatiFalesa teorema Universalnij slovnik enciklopediya 4 te vid K Teka 2006 nbsp Ce nezavershena stattya z matematiki Vi mozhete dopomogti proyektu vipravivshi abo dopisavshi yiyi Otrimano z https uk wikipedia org w index php title Teorema Falesa pro proporcijni vidrizki amp oldid 39559065
