Anybody understand Geometry?

yes they are. Ok picture a triangle. Ok. Now split it up into 3 parts. 2 are little tri. at the edge and the make 90 degree angles. Now the other piece in the middle has 3 sides to make up a box but the last piece instead of making it a squre comes to a point with another to form the top of the tri.
 
Hey all, yeah, I'm in Calc 1 this year so Geometry seems along ways in the past but I did like (wait for everyone to drop their jaws :jawdrop: "like and math in the same sent. NO WAY!") proofs (diclaimer-I liked them compared to the rest of the math I've ever done). So yeah I can't figure out where to even get a segment AE and DE in the problem...so if you could draw a little pic in paint and post it or if some how you could explain better, I'm sure I or someone else on here could help you out lots. :blink:
 
if ur doing AAS then-
FB is congruent to GC--that's in ur given
and angle FBA and angle GCD are 90* and congruent becauses that given
so there's an angle and a side...the other angle is giving me fits tho
 
aaahh...and then there is something about parallel lines and angles....something like the corresponding angles thm.....so like if you were to extend FB and GC they would be parallel lines...if you look on it on it's side (or maybe that's just helping me)...but then you have the corresponding angles and opposite angles thms. you can use to prove another angle
 
Princess Jeanie said:
aaahh...and then there is something about parallel lines and angles....something like the corresponding angles thm.....so like if you were to extend FB and GC they would be parallel lines...if you look on it on it's side (or maybe that's just helping me)...but then you have the corresponding angles and opposite angles thms. you can use to prove another angle
Wait...can u make a quick pic of what u reckon the triangle looks like, and then I can try and see it?
 
:D Ok, here is how I think this problems should work. (disclaimer...I'm occasionally wrong, altho this is a rare thing-unfortunatly it happens)

Ok, with the given info (which I'm not gonna restate) prove AB=DC

Statement ---> Proof (thm to back it up)

1. FB = GC ---> Given info
2. Angle FBA and angle GCD
are 90* and equal ---> Given info. (and I don't know you could
put something about perpendicular lines
and 90* angles)
3. Angle AFB = angle DGC ---> Properties of parallel lines...opposite angles
(I think, not corresponding angles)

So... 1=Side 2=Angle 3=Angle AAS

Because of the AAS thm you can say that they are congruent figures and then AB and DC are congruent parts to congruent figures which makes them equal......hope that helps ya. :cool:
 
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