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Let ABC be a triangle. We construct squares ABST and ACUV with centers O_1 and O_2, respectively, as shown. Let M be the midpoint of \overline{BC}.
unitsize(0.8 cm);pair A, B, C, M, S, T, U, V;pair[] O;A = (3,3);B = (0,0);C = (4,0);S = rotate(90,B)*(A);T = A + S - B;V = ro...
(a) Prove that \overline{BV} and \overline{CT} are equal in length and perpendicular.
(b) Prove that \overline{O_1 M} and \overline{O_2 M} are equal in length and perpendicular.
thanks
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hi,
I'm not getting a sensible diagram for this.
If As (3,3) B is (0,0) and C is (4,0) then (0,1) and (0,2) cannot be the centre of those squares.
Any chance you could post a picture please ?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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aha here it is:
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Yes Im a full member!!!!!! YEEESSSS!!!!
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Congratulations on becoming a full member.
Now for part a.
angle BAV = BAC + 90 = angle TAC
in triangles BAV and TAC that angle is equal, and you should be able to find two pairs of equal sides, so they are congruent, (SAS)
So TC = BV
Imagine rotating triangle TAC through 90 degrees, anticlockwise, around point A.
T rotates to B and C rotates to V, so one triangle becomes the other.
So TC must also rotate 90 to become BV, hence they are perpendicular.
part b
Consider triangles CVB and C02M.
M bisects BC and O2 bisects CV, so these triangles are similar and BV is parallel to MO2 and half its length
In the same way O1M is parallel to TC and half its length.
Part b follows directly from this.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Imagine rotating triangle TAC through 90 degrees, anticlockwise, around point A.
T rotates to B and C rotates to V, so one triangle becomes the other.
So TC must also rotate 90 to become BV, hence they are perpendicular.
Bob
How do we know that you have to rotate it 90 degrees?
Girls can be just as good as boys at math. We just need to get the same encouragement.
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Last edited by RandomPieKevin (2015-08-16 07:23:44)
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Can someone help with part b? Thanks.
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Can someone help me prove that
? Thanks.Offline
I thought I had done this in post 5.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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bob bundy wrote:Imagine rotating triangle TAC through 90 degrees, anticlockwise, around point A.
T rotates to B and C rotates to V, so one triangle becomes the other.
So TC must also rotate 90 to become BV, hence they are perpendicular.
Bob
How do we know that you have to rotate it 90 degrees?
Well, if we wanted to rotate point
anti-clockwise about point , it would end up at point This is because is thus must "travel" that angle to reach point .(Sorry if the LaTeX doesn't seem very neat, I tried to make it "proper" using the correct commands on this database. Quite ironic how LaTeX is supposed to make math look neater when in fact it makes it look quite messy on this website, if you ask me).
(Funny how I have edited this message like 20 times and I have only sat at my desktop for about eight minutes).
From,
ET ag
Last edited by ET ag (2017-02-24 10:32:54)
If a second was a minute, and a minute was an hour, how many hours would be in one day? (There are multiple answers and they are debatable)
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