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Bewyse en vermoedens in meetkunde

Jy het gesien hoe om meetkunde en die eienskappe van poligone te gebruik om die onbekende lengtes van sye en die groottes van hoeke van verskeie vierhoeke en poligone te vind. Ons gaan nou hierdie werk uitbrei om sommige van die eienskappe te bewys en probleme op te los. ʼn Vermoede is ʼn wiskundige se manier om te se: "Ek glo dit is waar, maar ek het geen bewys nie". Die volgende uitgewerkte voorbeelde sal help om dit duideliker te maak.

Gegee vierhoek ABCD, met A B C D en A D B C , bewys dat B A ^ D = B C ^ A en A B ^ C = A D ^ C .

  1. Ons maak die volgende skets en trek die diagonale.
  2. Gegee: A B C D en A D B C . Ons moet bewys A = C en B = D . In formele wiskundetaal sê ons dat ons gevra word om te bewys (RTP='requested to prove'): B A ^ D = B C ^ A en A B ^ C = A D ^ C .
  3. B A ^ C = A C ^ D ( verwisselende binnehoeke ) D A ^ C = B C ^ A ( verwisselende binnehoeke ) B A ^ D = B C ^ A
    Net so vind ons dat:
    A B ^ C = A D ^ C

In parallelogram ABCD, is die halveerlyne van die hoeke (AW, BX, CY en DZ) gekonstrueer:

Dit word ook gegee dat AB = CD , AD = BC , AB CD , AD BC , A ^ = C ^ , en B ^ = D ^ . Bewys dat MNOP parallelogram is.

  1. Gegee: AB = CD , AD = BC , AB CD , AD BC , A ^ = C ^ , and B ^ = D ^ . Bewys MNOP is ʼn parallelogram.
  2. In ADW and CBY D A ^ W = B C ^ Y ( gegee ) A D ^ C = A B ^ C ( gegee ) AD = BC (gegee) ADW = CBY (HHS) DW = BY
    In ABX and CDZ D C ^ Z = B A ^ X ( gegee ) Z D ^ C = X B ^ A ( gegee ) DC = AB (gegee) ABX CDZ (HHS) AX = CZ
    In XAM and ZCO X A ^ M = Z C ^ O ( gegee ) A X ^ M = C Z ^ O ( reeds bewys ) AX = CZ (reeds bewys) XAM COZ (HHS) A O ^ C = A M ^ X
    A M ^ X = P M ^ N (regoorstaande hoeke) C O ^ Z = N O ^ P (regoorstaande hoeke) P M ^ N = N O ^ P
    In BYN and DWP Y B ^ N = W D ^ P ( gegee ) B Y ^ N = W D ^ P ( reeds bewys ) DW = BY (reeds bewys) YBN WDP (AAS) B N ^ Y = D P ^ W
    D P ^ W = M P ^ O (regoorstaande hoeke) B N ^ Y = O N ^ M (regoorstaande hoeke) M P ^ O = O N ^ M

    MNOP is ʼn parallelogram (beide pare teenoorstaande 'e = , daarom is beide pare teenoorstaande sye ook parallel)

Dit is baie belangrik om daarop te let dat ʼn enkele teen-voorbeeld genoeg is om ʼn vermoede verkeerd te bewys. Selfs ʼn menigte ondersteunende voorbeelde is nog steeds geen bewys nie!

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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