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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).
 7 months ago
 7 months ago
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).
 7 months ago
 7 months ago

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mathstudent55Best ResponseYou've already chosen the best response.1
With coordinate geometry proofs, you need the coordinate plane.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
dw:1377488818630:dw
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
That is what you are proving right?
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
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 SideAngleSide. 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...
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
You need to label your drawing. Where are the points you are referring to?
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
dw:1377488962191:dw
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
dw:1377489047397:dw
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
dw:1377489011250:dw
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
The way you labeled it, D is not the midpoint of AB because D is not even on segment AB.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Okay so it would be a new point, C.
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
My understanding is that a coordinate geometry proof uses the coordinate plane, not just a drawing like regular geometry proofs use.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Okay...Well I cant draw a picture to answer it has to be in words only.
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
dw:1377489315487:dw
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Okay...Im a bit lost now...
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
dw:1377489369615:dw
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Why the new F point?
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
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.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
I mean! Since CD is the perpendicular bisector of AB, AE is congruent to BE?
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
Yes, but this needs to use the coordinates. That is what a coordinate proof is.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
So how would I do that, thats the part im not getting...
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
AE = a  0 = a = a BE = a  0 = a = a AE = BE = a
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
FE = d  0 = d = d
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Ohhhh...Okay makes sense!!!
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
We want to show that AF = BF. AF and BF are the hypotenuses of two right triangles, AEF and BEF, respectivly.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Right, So use the Pythagorean theorem with the coordinates?
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
Exactly. Pythagoras with coordinates.
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
\((AF)^2 = (AE)^2 + (FE)^2\) \( AF = \sqrt{ (AE)^2 + (FE)^2 } \) \( AF = \sqrt{ a^2 + d^2 } \)
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
Now we do the same for FB.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
\[(AF)^{2}=(BE)^{2}+(FE)^{2}\]
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
Yes, but we are doing BF now, not AF. \( (BF)^2=(BE)^2+(FE)^2 \) \( BF=\sqrt{(BE)^2+(FE)^2 } \) \( BF=\sqrt{a^2+d^2} \)
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Im a little confused on how you got from the 2nd to the 3rd...
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
We have shown that both AF and BF are equal to \( \sqrt{a^2 + d^2}\), so AF and BF are equal.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Is it because of what you said before?
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
You mean confused on going from (BE)^2 to a^2?
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
I was but then i read over it and got it!
 7 months ago

mathstudent55Best ResponseYou've already chosen the best response.1
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.
 7 months ago

OpenSessameBest ResponseYou've already chosen the best response.1
Okay thanks!! i have one more question, but i know the answer but not how to get their...
 7 months ago
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