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Mathematics 7 Online
OpenStudy (karatechopper):

Circumcenter is the point P. Need to find segments congruent to line segment CS.

OpenStudy (karatechopper):

OpenStudy (karatechopper):

Just need some assistance..please!

OpenStudy (amistre64):

congruent means "equal to" or equivalent

OpenStudy (anonymous):

the circumcenter is the point of concurrency of the three perpendicular bisectors (of the segments). The operative word here is bisector.

OpenStudy (karatechopper):

Yes. I know that,

OpenStudy (amistre64):

cant see the equation underneath so it difficult to see what the trianlge is spose to look like as opposed to its visual shape

OpenStudy (karatechopper):

Well its not an equation. They are just line segments

OpenStudy (amistre64):

its information pertaining to the shape that would help to define it otherwise it could just as well be an equilateral triangle

hartnn (hartnn):

since, PS is bisector, AS=CS

OpenStudy (anonymous):

since P is the circumcenter, it lies in the perpendicular BISECTOR BS

OpenStudy (karatechopper):

Yes, i have that segment down already. @hartnn

OpenStudy (karatechopper):

I am confused on that. You know i have to find congruent segments right? @r0n4ld

OpenStudy (amistre64):

at the moment, that is thhe only line segment that we can be sure of; the rest of the triangle is a mystery

OpenStudy (karatechopper):

Ok! Thats what i thought!

OpenStudy (karatechopper):

I was confused if there were any more thats why i posted

OpenStudy (karatechopper):

I have another question tho. Would line segment BT. be congruent to line segment BP?

hartnn (hartnn):

not at all, BP is hypotenuse of right triangle, whereas BT is the side.

hartnn (hartnn):

*BT is one of the leg.

OpenStudy (karatechopper):

Yes, but Point P is the circumcenter. So i would think every line segment is congruent.

OpenStudy (karatechopper):

If attached to Point P

hartnn (hartnn):

the 3 perpendicular bisectors will bisect 3 sides only... so u get 3 pirs of segments equal.

hartnn (hartnn):

*pairs

OpenStudy (karatechopper):

Would PS be one?

hartnn (hartnn):

nopes, because PS is not a part of a side of triangle.

hartnn (hartnn):

RA=RB SA=SC TB=TC

OpenStudy (karatechopper):

But they have to be congruent to BP

hartnn (hartnn):

there is a property : The perpendicular bisectors of the sides of a triangle meet at a point which is equidistant from the vertices so, BP=CP=AP also.

OpenStudy (karatechopper):

Oh..

OpenStudy (karatechopper):

BUT. P is the middle! So why wouldnt everyone thing attached to the P be congruent?

hartnn (hartnn):

u really can't say anything about PS,PT,PR

hartnn (hartnn):

'middle' doen;t mean middle of everything :P

hartnn (hartnn):

*doesn't

OpenStudy (karatechopper):

Awh man!

OpenStudy (karatechopper):

so its only CP and AP?

hartnn (hartnn):

yes...

hartnn (hartnn):

=BP

OpenStudy (karatechopper):

ok! THANKS!!!! :)

hartnn (hartnn):

welcome ^_^

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