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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).

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Ill draw it out?
With coordinate geometry proofs, you need the coordinate plane.

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Other answers:

That is what you are proving right?
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...
You need to label your drawing. Where are the points you are referring to?
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.
My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.
Okay...Well I cant draw a picture to answer it has to be in words only.
Okay...Im a bit lost now...
Why the new F point?
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.
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!!!
We want to show that AF = BF. AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.
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.
Yes, but we are doing BF now, not AF. \( (BF)^2=(BE)^2+(FE)^2 \) \( BF=\sqrt{(BE)^2+(FE)^2 } \) \( BF=\sqrt{a^2+d^2} \)
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!
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.
Okay thanks!! i have one more question, but i know the answer but not how to get their...
new post pls

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