Antwoord / Answer 1
AD = DC
. . . gegee / given
BE = EC
. . . DE halveer BC / DE bisects BC
DE || AB
. . . lyn deur middelpte. / line through midpoints
In ΔAFC
AD = CD
. . . gegee / given
DG || AF
. . . bewys hierbo / proven
CG = GF
. . . lyn || tweede sy / line || second side
Antwoord / Answer 2
2.1 In ΔABE
: AC = CE en / and CG || AB
. . . gegee / given
∴ BG = EG
. . . lyn deur midpt. || ander sy / line through midpt. || other side
2.2 In ΔBEF
:
GD || EF
. . . gegee / given
BG = GE
. . . bewys in 2.1 / proven 2.1
∴ BD = DF
. . . lyn deur midpt. || ander sy / line through midpt. || other side
2.3 In ΔABE
: AC = CE en / and CG || AB
. . . gegee / given
BG = GE
. . . bewys 2.1 / proven 2.1
CG = ½ AF
. . . mid. pt. stelling / midpoint theorem
Netso / Similarly
GD = ½ EF
∴ CG + GD = ½ AB + ½ EF
∴ CD = ½ (AB + EF)
Antwoord / Answer 3
3.1 In ΔPQR
: PT = TQ en / and TA || QR
. . . gegee / given
∴ PA = AR en / and TA = ½ QR
. . . lyn deur midpt. || ander sy /
line through midpt. || other side
In ΔPRS
: PA = AR (bewys / proven) en / and AU || PS
. . . gegee / given
∴ RU = US en / and AU = ½ PS
. . . lyn deur midpt. || ander sy /
line through midpt. || other side
In ΔSQR
: SB = BQ, SU = UR (bewys / proven) en / and BU || QR
. . . gegee / given
∴ BU = ½ QR
. . . midpt. stelling / midpoint theorem.
TA = ½ QR en / and BU = ½ QR
∴ TA = BU
3.2
AB = AU − BU
= ½ PS − ½ QR
= ½(PS − QR)
Antwoord / Answer 4
4.1 In ΔABD
: AK = KB en / and AN = ND
. . . gegee / given
∴ KN || BD en / and KN = ½ BD
. . . midpt. stelling / midpt. theorem
In ΔBDC
: BL = LC en / and DM = MC
. . . gegee / given
∴ LM || BD en / and LM = ½ BD
. . . midpt. stelling / midpt. theorem
∴ KN = en / and || LM;
4.2 KLMN is 'n parallelogram as KL || NM en KN || LM
KLMN is a parallelogram if KL || NM and KN || LM
Trek hoeklyn AC. / Draw diagonal AC.
In ΔABC
: AK = KB en / and BL = LC
. . . gegee / given
∴ KL || AC en / and KL = ½ AC
. . . midpt. stelling / midpt. theorem
In ΔADC
: AN = ND en / and DM = MC
. . . gegee / given
∴ NM || AC en / and NM = ½ AC
. . . midpt. stelling / midpt. theorem
∴ KL = en / and || NM;
KN || LM en / and KL || NM
∴ KLMN is 'n parallelogram / KLMN is a parallelogram
. . . pare oorstaande sye ewewydig / opposite pairs of sides are parallel
4.3 KL + LM + MN + NK = ½ AC + ½ BD + ½ AC + ½ BD
= AC + BD
Antwoord / Answer 5
∠K
1 = L
1 as DKML 'n parallelogram is (oorstaande hoeke gelyk) /
if DKML is a parallelogram (opposite angles equal)
Bewys dus dat DKML 'n parallelogram is [een paar oorstaande sye = en || ] /
Therefore prove that DKLM is a parallelogram by proving one pair of opposite sides = and ||.
In ΔDEF
: DK = KE en / and EM = MF
. . . gegee / given.
∴ KM || DF en / and KM = ½ DF
. . . middelpt. stelling / mid. point theorem.
DL = ½ DF
. . . L is die middelpunt van DF / L is the midpoint of DF.
∴ KM = DL en / and KM || DL
∴ DKML is 'n parallelogram (een paar oorstaande sye = en ||) /
DKML is a parallelogram (one pair of opposite sides = and ||)
∴ ∠K
1 = L
1 . . . oorstaande hoeke / opposite angles
// Wanneer sal ABCQ 'n parallelogram
// wees? Maak seker van die voorwaardes.
// 1. In die skets is MP || BC. Kan ons bewys
// dat AQ gelyk en ewewydig aan BC?
// Dan is ABCQ 'n parallelogram - een
// paar oorstaande sye gelyk en ewewydig.
// 2. In die skets is BP = PQ. Kan ons bewys
// dat AC en BQ mekaar halveer?
// As AP = PC en BP = PQ is ABCQ 'n
// parallelogram - hoeklyne halveer mekaar.
// When will ABCQ be a parallelogram? Make
// sure of the conditions???.
// 1. In the diagram MP || BC. Can we prove that
// AQ equal and parallel to BC?
// If so then ABCQ is a parallelogram - one
// pair of opposite sides equal an parallel.
// 2. In the diagram BP = PQ. Can we prove
// AC and BQ bisect one another?
// If AP = PC and BP = PQ then ABCQ is a
// parallelogram - diagonals bisect each other.
Bewys / Proof
In Δ ABC : AM = MB en / and MP || BC . . . gegee / given
∴ AP = PC en / and MP = ½ BC . . . lyn deur middelpunt van een sy en || tweede sy /
line through midpoint of one side and || second side
In Δ BAQ : AM = MB en / and BP = PQ . . . gegee / given
∴ MP || AQ en / and MP = ½ AQ . . . lyn deur middelpunt van een sy en || tweede sy /
line through midpoint of one side and || second side
∴ BC || MP || AQ en / and AQ = BC
∴ ABCQ is 'n parallelogram. . . . een paar oorst. sye = en ||
∴ ABCQ is a parallelogram . . . one pair of opp. sides = and ||
OF / OR
In Δ ABC : AM = MB en / and MP || BC . . . gegee / given
∴ AP = PC . . . lyn deur middelpunt van een sy en || tweede sy /
line through midpoint of one side and || second side
en / and BP = PQ . . . gegee / given
∴ AC en BQ halveer mekaar. / AC and BQ bisect one another.
∴ ABCQ is 'n parallelogram. . . . hoeklyne halveer mekaar.
∴ ABCQ is a parallelogram . . . diagonals bisect one another.
Antwoord / Answer 7
In Δ EMF
: K is die middelpunt van EF en EM || KN
. . . gegee
/ K is the midpoint of EF and EM || KN
. . . given
∴ MN = NF
. . . lyn deur middelpunt van een sy en || tweede sy /
line through midpoint of one side and || second side
In Δ DKN
: L is die middelpunt van DK en LM || KN
. . . gegee
/ L is the midpoint of DK and LM || KN
. . . given
∴ DM = MN
. . . lyn deur middelpunt van een sy en || tweede sy
/ line through midpoint of one side and || second side
∴ DM = MN = NF
∴ DF = DM + MN + NF
= 3 DM