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- anonymous

Geometry :D
Suppose f and g are two isometries such that f(A) = g(A), f(B) = g(B), and f(C) = g(C) for some nondegenerate triangle ΔABC. Show that f = g

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- anonymous

- chestercat

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- anonymous

I feel like this should be easy, but still getting used to all the geometry theorems and what not.

- anonymous

@freckles know anything about this stuff?

- anonymous

@ganeshie8 Know any of this stuff?

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- ganeshie8

idk much but i found this
#3
http://www.ms.uky.edu/~droyster/courses/spring04/homework/restricted/Homework2%20Solns.pdf
@ikram002p did non eucledean geometries a couple of years ago

- anonymous

Ah, nice find! I found something else online, but it used a theorem that I couldnt use based on the sequence of my text. But alright, if ikram knows some of this stuff, Ill see what he can do when I have questions. I think I can just use what that link gives. Thanks :)

- ganeshie8

good luck! :)

- ikram002p

hey first u need to know what isometries functions means, it means a rigid transforms function which maps a shape to some where else but without changing size in other way both origin and transforms are congruent.

- ikram002p

lets do it in geometry style
|dw:1433400064930:dw|

- ikram002p

so since both are isometries we have 3 congruent triangles, now here is a thing why it called rigid transformation it since u do not change it size and shape , like triangle do not become a circle for example or dont become another triangle with same size but different sides that cannot happen.

- ikram002p

now step two, it says if
f(A) = g(A), f(B) = g(B), and f(C) = g(C), then show Show that f = g
in euclidean geometry its ok if u wanna only graph it that works like this it would be a proof by itself
|dw:1433400511306:dw|

- ikram002p

now since you wanna it theoretically
f(A)=g(A)
g^-1o f(A)=g^_1 o g(A) =A
so if u apply the function g^-1 o f on the other two points u got
g^-1 o f(B)=B
g^-1 o f(C)=C
according to a theorem (idk if ts given in ur book or u need to prove it let me know in both cases )
g^-1o f must be the identity which means :-
g^-1 o f(P)=P , for any P on the triangle
g o g^-1 o f(P)= g(P)
f(P)=g(P)
which is done :D
let me know if something is not clear enough, good luck !!

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