Is anyone good with proofs in mathematics?
Help me with math please .-.
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StevenNL2000 I heard you're good with proofs, can you help me with these two problems?
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<HAB is congruent to <WAX by vert. < theorem
<HBA is congruent to <AWX by alt. interior < theorem, same for <AHB and <XWA
by the aaa similarity thm. the triangles are similarit's been a while since i've taken geometry so idk about the 2nd one but i think the 1st one is right
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Thanks bro, could you take care of the last one?
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Oh. Thanks.
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CodexTheDuplex I've fixed your image embed
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@lyicx#6000
Thanks, could you help me with the second proof? -
@CodexTheDuplex#6002 that looks like some form of trigonometry or something. been awhile since i've done math so i dont know if ill be able to help enough
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volleo6144 is great at maths, I'm sure he knows
@CodexTheDuplex#6002 -
I haven't done this kind of bloody maths since my JC, will be revisiting it this year or next year.
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Tozzit hey heard youre good at math
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CodexTheDuplex im going to repost your image in here. please dont make another thread for it in deep discussion with a single image.
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okay thank you.
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shrimp any ideas?
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can someone solve the second proof please
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First image, proof #1:
∠BHX = ∠HXW because they are alternate angles on the parallel lines HB and XW.
It is easy to see that ∠BHX = ∠BHA and ∠HXW = ∠AXW, so ∠BHA = ∠AXW.
∠HBW = ∠BWX because they are alternate angles on the parallel lines HB and XW.
It is easy to see that ∠HBW = ∠HBA and ∠BWX = ∠AWX, so ∠HBA = ∠AWX.
Finally, ∠BAH = ∠WAX because they are opposite angles of two intersecting lines.
We have shown that all three pairs of angles are equal, so by the AAA triangle similarity test, △HAB ~ △XAW.First image, proof #2:
△PFY has sides FY = 16 and FP = 12.
FK = FY+YK = 16+12 = 28
FS = FP+PS = 12+9 = 21
So △SFK has sides FK = 28 and FS = 21.
FK/FY = 28/16 = 1.75, so the proportion of FY to FK is 1:1.75.
FS/FP = 21/12 = 1.75, so the proportion of FS to FP is 1:1.75.
It is easy to see that ∠SFK = ∠PFY.
We have shown that there are two pairs of sides with equal proportions and that the included angle is equal, so by the SAS triangle similarity test, △SFK ~ △PFY.Second image:
△AZY has sides AZ and AY.
AZ = ZB, so AB = AZ+ZB = AZ+AZ = 2AZ.
AY = YC, so AC = AY+YC = AY+AY = 2AY.
So △ABC has sides AB = 2AZ and AC = 2AY.
AB/AZ = 2AZ/AZ = 2, so the proportion of AZ to AB is 1:2.
AC/AY = 2AY/AY = 2, so the proportion of AY to AC is 1:2.
It is easy to see that ∠BAC = ∠ZAY.
We have shown that there are two pairs of sides with equal proportions and that the included angle is equal, so by the SAS triangle similarity test, △ABC ~ △AZY.
Then the proportion of the third side has to be 1:2 as well, so BC = 2ZY.
BC = 2ZY
3x+6 = 2(x+6)
3x+6 = 2x+12
x+6 = 12
x = 6 -
@neo#6007 got here too late smh
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why is steven good at everything smh
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@Tozzit#6156 hey heard youre good at math
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wild1145
July 16, 2022 at 11:38 PM Moved the thread from forum Imported from Flarum to forum General Discussion.