Today was our final, and I finished it, thinking I did mighty fine.

 

 

 

But there was one question that annoyed me, it was multiple choice.

 

 

 

It showed this figure:

 

(not to scale, added numbers for angles for convience)

 

 

and it stated that seg.BC is 12 units.

 

 

 

Then it said to name anouther measurement, two of the choices were:

 

"seg.DC is 12 units" (right answer) and the other was "seg.AB is 12 units".

 

The obviously easyier answer was the first, but couldn't you prove the second one also?

 

 

 

So that's what I tried doing after the final.

 

 

 

I got pretty far, but not far enough. This is what I was able to prove up to:

 

 

(Tell me if I'm wrong about something)

 

 

 

I need to be able to prove atleast one of a few things:

 

• ang.3 congruent to ang. 4 or 10
   
  and.3 congruent to ang. 10 or 4
   
  ang.1 or 2 congruent to ang.12 or 11
   
  seg.AE congruent seg.EC
   
  seg.AB parallel seg.DC
   
  seg.AB congruent sec.DC
   
  seg.AD parallel seg.BC
   
  seg.AD congruent seg.BC
   
  seg.DC perpendicular bisector to seg.AC

 

 

 

any help?

 

 

 

(Use any theorems you can, and you can draw new lines, make new planes and us a foot, whatever you have to)

So angle E and angle 9 aren't the same?

 

 

 

If they aren't then the bottom and top lines won't be the same.

E is the name of the point, you can't call it angle E because that can refer to many angles.

Make another angle by extending line AB out beyond B, and call that endpoint Z.

 

 

 

Now since angle DBZ is supplemetary to angle 3 and also to angle 10, angle 10 is congruent to angle 3.

 

 

 

That should be able to prove it.

 

 

 

EDIT: Oh, right, you can't prove those are parallel lines. . .

 

 

 

I don't think it's possible.

Make another angle by extending line AB out beyond B, and call that endpoint Z.

 

 

 

Now since angle DBZ is supplemetary to angle 3 and also to angle 10, angle 10 is congruent to angle 3.

 

 

 

That should be able to prove it.

 

 

 

How would you say that angle 10 is supp. to DBZ? I haven't proved parallel lines.

 

 

 

 

 

I was thinking...

 

 

 

Is the converse of "If parallelogram, then diagonals form congruent triangles" true? (If diagonals form congruent triangles, then parallelogram)

 

 

 

And would just one set of congruent triangles be enough? (DAC and BAC)

 

 

 

 

 

I'm also trying to figure out if using negetive statements and proving them wrong could help...

 

 

 

ex:

 

angle 3 is either congruent or not congruent to angle 10

 

if not congruent then... *disprove given, or proven statement*; therefor, congruent.

Well, because angle 6 is 90ÃÆââ¬Å¡Ãâð, angle 7,8, and 9 are, too. Which means that AB = BC = CD = DA (it's a rhombus). I'm sure that this could be put into a proof, but I'm tired and I was never very good at triangle proofs. :P

 

 

 

But if you didn't see that, because angle 7 = angle 8 (they're vertical angles), side AB = DC (equal angles form equal sides), so either of the two choices would be right.

Well, because angle 6 is 90ÃÆââ¬Å¡Ãâð, angle 7,8, and 9 are, too. Which means that AB = BC = CD = DA (it's a rhombus). I'm sure that this could be put into a proof, but I'm tired and I was never very good at triangle proofs. :P

 

 

 

But if you didn't see that, because angle 7 = angle 8 (they're vertical angles), side AB = DC (equal angles form equal sides), so either of the two choices would be right.

 

 

 

That's only true if the sides extending from the angle are congruent (therefore side-angle-side)

 

 

 

Here's a picture that proves what you said is wrong:

 

 

 

Well, because angle 6 is 90ÃÆââ¬Å¡Ãâð, angle 7,8, and 9 are, too. Which means that AB = BC = CD = DA (it's a rhombus). I'm sure that this could be put into a proof, but I'm tired and I was never very good at triangle proofs. :P

 

 

 

But if you didn't see that, because angle 7 = angle 8 (they're vertical angles), side AB = DC (equal angles form equal sides), so either of the two choices would be right.

 

 

 

Whoa, there, back up. You can't go saying it's a rhombus when it hasn't even been proved to be a parallelogram.

 

 

 

And since I don't see a way to prove it's a parallelogram, it probably can't be proven that Ab is congruent to BC.

All the info we've got is that angle 6 (and thus 7, 8 and 9 as well) and that DE and EB are of the same length, right?

 

We can't know if AB and BC are congruent, or whether their angles are the same. The only thing we can know is that AD and AB are congruent.

 

All of these possible figures meet the requirements.

 

I don't see a way to prove it. With what you have to work with, you can't find the lengths of AB, AE, or AD because you have no idea how far A is from the center.

would love to try but i gotta go, need all the sleep i can get before i fail my geo exam tomarrow.

 

 

 

honestly, im switching to basic...

The shape is a kite. The two "diagonals" (don't know the exact term) are perpendicular, and one diagonal bisects the other.

 

 

 

 

 

 

 

Or you could say that triangle BEC is congruent to triangle DEC by SAS.

 

segment BE is congruent to segment DE

 

Angle 8 is congruent to Angle 9

 

segment EC is congruent to segment EC

 

 

 

 

 

Therefore, DC = BC