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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I feel like this should be easy, but still getting used to all the geometry theorems and what not.
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idk much but i found this
@ikram002p did non eucledean geometries a couple of years ago
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 :)
good luck! :)
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.
lets do it in geometry style
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.
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
now since you wanna it theoretically
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)
which is done :D
let me know if something is not clear enough, good luck !!