Geometry Proofs (2 Viewers)

Smile12345

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I'm really not sure about how to do this one... Could someone help me please?

Diagram: Q6.png

Given Info: AB and BC are straight lines. XBZY is a parallelogram. ZC = ZY and AX = BX. The points A, Y and C are collinear.

a) Prove that Triangle AXY is congruent to Triangle CYZ... - What proof should I use?
b) Prove that XBZY is a rhombus....

Thanks for all your help in advance. :)
 
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Carrotsticks

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I'm really not sure about how to do this one... Could someone help me please?

Diagram: View attachment 28361

Given Info: AB and BC are straight lines. XBZY is a parallelogram. ZC = ZY and AX = BX. The points A, Y and C are collinear.

a) Prove that Triangle AZY is congruent to Triangle CYZ... - What proof should I use?
b) Prove that XBZY is a rhombus....

Thanks for all your help in advance. :)
Can you please confirm that (a) was typed correctly?
 

Smile12345

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Can you please confirm that (a) was typed correctly?
Sorry a) was typed incorrectly... Please forgive me... :):) I've changed it now ( It should be: Prove that Triangle AXY is congruent to Triangle CYZ...)
 

Carrotsticks

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Oops, the green kinda came out funky. But it's meant to be 180-2x.
 

Smile12345

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Sorry, unfortunately I can't see your pic... :( Is it possible to put it into another program?

Sorry, I hope you haven't taken that Triangle ABC is a right angle... My diagram isn't quite right... My bad fault... Please forgive me.
 
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Smile12345

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It's more like this but the smaller two triangles are congruent...Q6.png
 

Carrotsticks

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Sorry, unfortunately I can't see your pic... :( Is it possible to put it into another program?

Sorry, I hope you haven't taken that Triangle ABC is a right angle... My diagram isn't quite right... My bad fault... Please forgive me.
I didn't assume it was right angled.

Capture.jpg
 

Drongoski

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Q6: My version


Since XBZY is a //gram: XY//BC and AB//YZ

ZY = XB (.: = AX) [opp sides of //gram]

< AXY = < ABC = @ say [corresp <'s, XY//BZ]

< YZC = < ABC = @ [corresp <'s, AB//YZ]

< AYX = < YCZ = & say [corresp <'s, XY//BC]

.: in triangles AXY and YZC:

- < AYX = < YZC [=@]

- < AYX = < YCZ [=&]

- AX = YZ

.: triangles AXY and YZC are congruent [AAS]

.: their corresponding sides are equal

.: XY = ZC [= ZY]

.: XY = XB [since ZY = XB]

.: //gram XBZY is a rhombus [since its 4 sides are equal]
 
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Smile12345

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Q6: My version


Since XBZY is a //gram: XY//BC and AB//YZ

ZY = XB (.: = AX) [opp sides of //gram]

< AXY = < ABC = @ say [corresp <'s, XY//BZ]

<YZC = < ABC = @ [corresp <'s, AB//YZ]

<AYX = < YCZ = & say [corresp <'s, XY//BC]

.: in triangles AXY and YZC:

- <AYX = < YZC [=@]

- < AYX = < YCZ [=&]

- AX = YZ

.: triangles AXY and YZC are congruent [AAS]

.: their corresponding sides are equal

.: XY = ZC [= ZY]

.: XY = XB [since ZY = XB]

.: //gram XBZY is a rhombus [since its 4 sides are equal]
Thanks heaps... :)
 

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