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Unformatted text preview: Two Proofs Involving Isometries Theorem: The inverse of an isometry is an isometry.
Proof: Suppose that a : P -+ P is an isometry.
Let B = (1‘1 . Then, for any point X in P, [3(X) =X ifandonlyif a(X') = X, and, inthiscase, a°B(X) = X, since a°B(X) = a(B(X) ) = a(X ) = X.
Let X and Y be any two points in the plane P. [Weneedtoshowthat ( [3(X) B(Y)) = (XY).] Let x'= B(X) andlet Y'= B(Y). Then, a(X') =X and a(Y') =Y. V Sincea isanisometry, (a(X')a(Y')) = X'Y . V ' Thus,XY = X'Y', so, X Y = XY.
But, X'= B(X) andlet Y'= B(Y). Thus, (50‘) [3(Y)) = (XY) QED Problem: It is given that: __ ._ 6—)
AB is the diameter of circle C( Z , ZB ) . M is the midpoint of ZB and line s = AB . Line t is perpendicular to line 5 at point M . Line t intersects C(Z , ZB)at points H and K. Line t ToProve: (1)Rt(Z)=B and Rt(B)=Z. (2)RS(H)=K and RS(K)=H. Lines (3) 115(3) = B and Rs° RS(H) = H. Proof: (1) Since line t is the perpendicular bisector of segment ZB- , R t ( Z) = B and Rt ( B ) = Z , by deﬁnition of a reﬂection about the mirror line t . 9—) (—) .— ..._....
(2) Since line s = AB and line t = HK , diamter AB is perpendicular to chord HK. By Theorem 4.5.4 ("If a diameter is perpendicular to a chord, then the diameter bisects it"), diameter AB bisects chord Thus, line 5 is the perpendicular bisector of R . Therefore, R s (H ) = K and R s ( K) = H , by deﬁnition of a reﬂection about the mrror line s .
(3) Since B is a point on line s , R s ( B ) = B , by deﬁnition of a reﬂection about the mrror line 5 .
By deﬁnition of composition of functions, R 5 ° R s (H ) = R s (R s ( H ) ) .
Asshownin(2), RS(H) = K. Therefore, Rs° RS(H) = RS(RS(H)) = RS(K) = H. Therefore, RS° Rs(H) = H. QED ...
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- Spring '10
- Elementary geometry, 5 °, isometries, rs° rs