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Posted

I'm having trouble figuring this out right off hand, so I figured I'd see if anyone could explain it, or give some insight that I could use on making a two-column proof for this. Here's the info:

Picture14-8.png

Posted

Quite easy:

I dont know how you proove, but here is the proof

Before that take the point of intersection of the line CG and EA as.. Say point M

In ABDE is a //gram, CD is // to EM and DF is // to CG

In CDFG is a //gram, CD is // to FG and DE is // to BA

Therefore, EM is // to FG and CM is // to BA (The Trick and it means that FGCM is a //gram too)

Now Angle G = Angle M (FG is // to EM)

Now Angle M = Angle A (CM is // to BA)

Hence Angle G = Angle A.

Posted

Eh... what do you mean by two column proof?

Well if you just want explanations, i can give a fairly detailed one:

First part

Given: ABDE and CDFG are parallelograms --> BD//AE (BD parallel to AE *from diagram*) and CD//FG

Since C lies on line BD, BD//CD

From the two statements above, we know that BD//AE//CD//GF --(1)

Second part

Given: ABDE and CDFG are parallelograms --> DF//CG and BA//DE from diagram

Since E lies on DF, DE//DF

From these two statements above, we know that DE//DF//CG//BA --(2)

We need to prove: angle G is the same as angle A,

which means this -

PROVE angle CGF = angle BAE

from equation (2), we know that CG//BA

from equation (1), we know that GF//AE

Thus, using rule of corresponding angles

CGF ~= BAE

Posted

If ABDE and CDFG are parallelograms, you could do it like this:

1) ABDE and CDFG are parallelograms. (given)

2) Angle A is congruent to Angle D. (opposite angels of a parallelogram are congruent)

3) Angel D is congruent to Angle G. (same as 2)

4) Angel A is congruent to Angle G. (If two angles are congruent to the same angle, they are congruent)

Posted

*meh* YOU GUYS STOLE MY EXPLANATION!!!! Sorry... I'm having a hard time getting onto DCW other than school hours and five minutes in the morning... Therefore I can't post very long things.

  • 2 weeks later...

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