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Ill draw it out?

With coordinate geometry proofs, you need the coordinate plane.

|dw:1377488818630:dw|

That is what you are proving right?

You need to label your drawing. Where are the points you are referring to?

|dw:1377488962191:dw|

|dw:1377489047397:dw|

|dw:1377489011250:dw|

The way you labeled it, D is not the midpoint of AB because D is not even on segment AB.

Okay so it would be a new point, C.

Okay...Well I cant draw a picture to answer it has to be in words only.

|dw:1377489315487:dw|

Okay...Im a bit lost now...

|dw:1377489369615:dw|

Why the new F point?

I mean!
Since CD is the perpendicular bisector of AB, AE is congruent to BE?

Yes, but this needs to use the coordinates. That is what a coordinate proof is.

So how would I do that, thats the part im not getting...

AE = |-a - 0| = |-a| = a
BE = |a - 0| = |a| = a
AE = BE = a

FE = |d - 0| = |d| = d

Ohhhh...Okay makes sense!!!

Right, So use the Pythagorean theorem with the coordinates?

Exactly. Pythagoras with coordinates.

\((AF)^2 = (AE)^2 + (FE)^2\)
\( AF = \sqrt{ (AE)^2 + (FE)^2 } \)
\( AF = \sqrt{ a^2 + d^2 } \)

Now we do the same for FB.

Okay...

\[(AF)^{2}=(BE)^{2}+(FE)^{2}\]

Im a little confused on how you got from the 2nd to the 3rd...

We have shown that both AF and BF are equal to \( \sqrt{a^2 + d^2}\), so AF and BF are equal.

Is it because of what you said before?

You mean confused on going from (BE)^2 to a^2?

I was but then i read over it and got it!

Okay thanks!! i have one more question, but i know the answer but not how to get their...

new post pls