Geometry Theorems (3U) (1 Viewer)

[テリー]

I'm havin abita problems with the new 3u Geometry Theorems
Theorem 1:
If the intercepts cut off on one transversal by three (or more) parallel lines are equal, the intercepts cut off by them on any other transversal are also equal.
Given: AB=BC,
PROVE: BD = EF

 

[テリー]

Theorem 2:
The straight line joining the mid-points of two sides of a triangle is:
a.) parallel to the third side,
b.) one half of the length of the third side.
Given: D and E are mid-points of AB and AC
PROVE: DB||BD & DE = 1/2 of BC

 

[SoulSearcher]

Hmm this first question should go like this, assuming you have to prove DE = EF.

allow the angle sign to be <, I cant remember how to do the actual one
we have to prove that ∆DEG is congruent to ∆EHF

<ABG + <BAD = 180 (cointerior angles)
AB = DG (given), and
AD || BG, therefore quadrilateral ABGD is a parallelogram.
<BCH + <CBE = 180 (cointerior angles)
BC = EH (given), and
BG || CH, therefore quadrilateral BCHE is a parallelogram.

now
<ABG = <ADG (opposite angles of a parallelogram)
<ADG = <DGE (alternate angles)
therefore <ABG = <DGE
now
<ABG = <BCH (corresponding angles)
<BCH = <BEH (opposite angles of a parallelogram)
<BEH = <EHF (alternate angles)
therefore <ABG = <EHF
therefore <DGE = <EHF (angle)

now
<EDX = <DEG (alternate angles, X is a point to the right of the point D on the line AD) and
<EDX = <FEY (corresponding angles, Y is a point to the right of E on the line BE) and
<FEY = <EFH (alternate angles)
therefore <EFH = <EDX and thus
<DEG = <EFH (angle)
since DG = EH (given, side)
∆DGE is congruent to ∆EHF (AAS)
and therefore DE = EF (corresponding sides of congruent triangles)

 

[GaDaMIt]

Hmm this first question should go like this, assuming you have to prove DE = EF.

allow the angle sign to be <, I cant remember how to do the actual one
we have to prove that ∆DEG is congruent to ∆EHF

<ABG + <BAD = 180 (cointerior angles)
AB = DG (given), and
AD || BG, therefore quadrilateral ABGD is a parallelogram.
<BCH + <CBE = 180 (cointerior angles)
BC = EH (given), and
BG || CH, therefore quadrilateral BCHE is a parallelogram.

now
<ABG = <ADG (opposite angles of a parallelogram)
<ADG = <DGE (alternate angles)
therefore <ABG = <DGE
now
<ABG = <BCH (corresponding angles)
<BCH = <BEH (opposite angles of a parallelogram)
<BEH = <EHF (alternate angles)
therefore <ABG = <EHF
therefore <DGE = <EHF (angle)

now
<EDX = <DEG (alternate angles, X is a point to the right of the point D on the line AD) and
<EDX = <FEY (corresponding angles, Y is a point to the right of E on the line BE) and
<FEY = <EFH (alternate angles)
therefore <EFH = <EDX and thus
<DEG = <EFH (angle)
since DG = EH (given, side)
∆DGE is congruent to ∆EHF (AAS)
and therefore DE = EF (corresponding sides of congruent triangles)

^^ Word

 

[SoulSearcher]

For the second question, a couple of things need to be shown, that <ADE = <ABC and that ∆ADE is similar to ∆ABC

now
<ADE = <AED (isosceles triangle ∆ADE has equal base angles)
therefore <DAE + 2<ADE = 180
i.e. that <DAE = 180 - 2<ADE
now
<ABC = <ACB (isosceles triangle ∆ABC has equal base angles)
therefore <DAE + 2<ABC = 180
i.e. that <DAE = 180 - 2<ABC
therefore 180 - 2<ADE = 180 - 2<ABC
therefore 2<ADE = 2<ABC
<ADE = <ABC
therefore <ADE and <ABC are corresponding angles, and therefore DE || BC

now <BAC = <DAE (common)
and <ADE = <ABC (shown above)
so ∆ABC ||| ∆ADE
therefore DE / BC = AD / AB
but 2AD = AB (midpoint of AB makes AD = DB, which is given)
i.e. that AD / AB = 1 / 2
therefore DE / BC = 1 / 2
therefore DE = 1/2 * BC

 

[SoulSearcher]

^^ Word

lol.

 

[Pubert]

I don't think these questions are three unit, they look like yr 10 advanced work.

 

[テリー]

Thx SoulSearch
and this IS 3U maths... its in the 2U book nd it says for 3U students only...




___________________________
3U Maths
ESL
Physics
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Im a typical Honkie

 

[Wooz]

What book, fitzpatrick?

 

[SoulSearcher]

Probably, I've seen it somewhere in the Fitzpatrick 2 unit book.

 

[テリー]

yea, its from that book nd it suk bad

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