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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.

Exercise 1: Bewyse - 1

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

Solution

  1. Step 1. Trek ʼn diagram: Ons maak die volgende skets en trek die diagonale.
    Figure 1
    Figure 1 (geomproof1.png)
  2. Step 2. Skryf neer wat is gegee en wat word gevra: Gegee: ABCDABCD en ADBCADBC. Ons moet bewys A=CA=C en B=DB=D. In formele wiskundetaal sê ons dat ons gevra word om te bewys (RTP='requested to prove'): B A ^ D = B C ^ A B A ^ D= B C ^ A en A B ^ C = A D ^ C A B ^ C= A D ^ C.
  3. Step 3. Los die probleem op:
    B A ^ C = A C ^ D ( verwisselende binnehoeke ) D A ^ C = B C ^ A ( verwisselende binnehoeke ) B A ^ D = B C ^ A B A ^ C = A C ^ D ( verwisselende binnehoeke ) D A ^ C = B C ^ A ( verwisselende binnehoeke ) B A ^ D = B C ^ A
    (1)
    Net so vind ons dat:
    A B ^ C = A D ^ C A B ^ C= A D ^ C
    (2)

Proofs - 2 2

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

Figure 2
Figure 2 (geomproof2.png)
Dit word ook gegee dat AB=CDAB=CD, AD=BCAD=BC, ABCDABCD, ADBCADBC, A ^ = C ^ A ^ = C ^ , en B ^ = D ^ B ^ = D ^ . Bewys dat MNOP parallelogram is.

Solution

  1. Step 1. Skryf neer wat gegee is en wat jy gevra word om te bewys: Gegee: AB=CDAB=CD, AD=BCAD=BC, ABCDABCD, ADBCADBC, A ^ = C ^ A ^ = C ^ , and B ^ = D ^ B ^ = D ^ . Bewys MNOP is ʼn parallelogram.
  2. Step 2. Los die probleem op:
    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 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
    (3)
    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 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
    (4)
    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 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
    (5)
    A M ^ X = P M ^ N(regoorstaande hoeke) C O ^ Z = N O ^ P(regoorstaande hoeke) P M ^ N = N O ^ P A M ^ X = P M ^ N(regoorstaande hoeke) C O ^ Z = N O ^ P(regoorstaande hoeke) P M ^ N = N O ^ P
    (6)
    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 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
    (7)
    D P ^ W = M P ^ O(regoorstaande hoeke) B N ^ Y = O N ^ M(regoorstaande hoeke) M P ^ O = O N ^ M D P ^ W = M P ^ O(regoorstaande hoeke) B N ^ Y = O N ^ M(regoorstaande hoeke) M P ^ O = O N ^ M
    (8)

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

Warning:

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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