Prove using coordinate geometry: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
You are given a line segment (AB) and a perpendicular line segment (CD).

- anonymous

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- anonymous

Ill draw it out?

- mathstudent55

With coordinate geometry proofs, you need the coordinate plane.

- anonymous

|dw:1377488818630:dw|

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## More answers

- anonymous

That is what you are proving right?

- anonymous

In the triangle ADC and BDC, AD and BD are congruent because D is the midpoint of the segment AB. Also the side DC is common in both the triangles. This mean that angle CDA and angle CDB are both 90 degrees meaning they are congruent. This means the triangles are congruent because of the postulate Side-Angle-Side. So since the triangles are congruent, the distance from the two endpoints have to be equal.
^ that is what i said first but its wrong...

- mathstudent55

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

- anonymous

|dw:1377488962191:dw|

- anonymous

|dw:1377489047397:dw|

- mathstudent55

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- mathstudent55

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

- anonymous

Okay so it would be a new point, C.

- mathstudent55

My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.

- anonymous

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

- mathstudent55

|dw:1377489315487:dw|

- anonymous

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

- mathstudent55

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- anonymous

Why the new F point?

- mathstudent55

Given: Segment CD is the perpendicular bisector of segment AB.
Prove: F, a point on the perpendicular bisector of AB, is equidistant from A and B.

- anonymous

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

- mathstudent55

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

- anonymous

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

- mathstudent55

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

- mathstudent55

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

- anonymous

Ohhhh...Okay makes sense!!!

- mathstudent55

We want to show that AF = BF.
AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.

- anonymous

Right, So use the Pythagorean theorem with the coordinates?

- mathstudent55

Exactly. Pythagoras with coordinates.

- mathstudent55

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

- mathstudent55

Now we do the same for FB.

- anonymous

Okay...

- anonymous

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

- mathstudent55

Yes, but we are doing BF now, not AF.
\( (BF)^2=(BE)^2+(FE)^2 \)
\( BF=\sqrt{(BE)^2+(FE)^2 } \)
\( BF=\sqrt{a^2+d^2} \)

- anonymous

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

- mathstudent55

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

- anonymous

Is it because of what you said before?

- mathstudent55

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

- anonymous

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

- mathstudent55

It's bec of these statements above:
AE = |-a - 0| = |-a| = a
BE = |a - 0| = |a| = a
AE = BE = a
FE = |d - 0| = |d| = d
Oh, I see that you understand already. Great.

- anonymous

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

- mathstudent55

new post pls

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