-?_ capm 7G=! 8L@78A=I70T%HH!s  lThree Points Theorem,I3zA?0><80 an @7 C=7 C=C!O! @@U 8L@7U JU Hv  g7X|}9:jFProofs: (1) Copy and print the sketch provided below and carry out a paper-and-pencil proof using a compass and straightedge to construct the image T(X) = X' where T is the isometry that maps triangle ABC onto A'B'C'. (2) Follow these directions with Sketchpad to construct your own triangle ABC and some congruent image of it in the plane. Then proceed with the demonstration proof, outlined below. Place 3 noncollinear points A, B, C. Choose a point A' in the plane to be T(A); choose B'= T(B) on the circle with center A' and radius AB (since T preserves distances, we must have |AB| = |A'B'|; then choose C'= T(C) to be one of the two intersection points of the circle centered at A' with radius |AC| and the circle centered at B' with radius |BC| (why must C' be chosen this way?). Now place a point X (somewhere near triangle ABC to make the picture easier to see). (3) Use the sketch provided below and follow the directions in the demonstration proof to construct the unique point T(X) = X' where T is the isometry that maps triangle ABC onto A'B'C'. Demonstration proof (carry out with Sketchpad): There is only one point X' such that the distances X'A', X'B', X'C' are equal to the corresponding distances XA, XB, XC, so this point must be chosen as T(X). The point X' is constructed as the unique intersection point of three circles: the circle centered at A' with radius AX, the circle centered at B' with radius BX, and the circle centered at C' with radius CX. You will need to construct the three segments AX, BX, CX in order to construct the three circles with Sketchpad (use the Construct menu, Circle by Center + Radius). When you have done the construction, drag the position of X to see what happens. that the distances X'A', X'B' A', X'B' A', X'B'0distances X'A', X'B' A', X'B' A', X'Fn 8L@7+`,`IlJllAlB `The images of three noncollinear points completely determine a plane isometry. That is, if the images T(A) = A', T(B) = B', T(C) = C' under an isometry T are known, then the image T(X) of every other point X in the plane is determined by the condition that all distances are preserved.l distances are preserved. this sketch, select, in order (hold shift), shaded interior, points A, B,Pres8L@7AT( #H8~0~T wC H AFRe8L@7BT( #H8~0~T wB EJD 8L@7CT( #H8~0~T wCSB =B8L@7DT( #H8~0~T wC²8L@7ET( #H8~0~T wCgC+;0@58L@7FT( #H8~0~T wCBt.r3w8L@7XT"@ H0@H( @@T lTdBB@@J 8L@7pGPbdStyleGx-IB Use Subscript@ @p@uCSBB?J 8L@7o\st ![\st!#Y[tv#C HCSB?@  8L@7n""""""""""""""""""UUUUJHBC H?B 8L@7jT( #H8~0~T wC²CgC+?f2k78L@7CT( #H8~0~T wCBh~8L@7BT( #H8~0~T wCNB627U8L@7AT( #H8~0~T wCۍBOZ_I8L@7CT( #H8~0~T wCyB uzI8L@7BT( #H8~0~T wCFC`~ I8L@7AT( #H8~0~T wCEC  YzH 8L@7sHHHlCFC`~CyB?tY 8L@7r ![\st CECCFC`~?Y 8L@7q8J,A@@AACyBCEC?afI8L@7C'T"@ H0@G @@T lTdC`A!INI8L@7BT"@ H0@G @@T lTdCmC42 HfI 8L@7sHHHlCmC42C`A!?HI 8L@7rծ ![\st CECCmC42?`I 8L@7q8J,A@@AAC`A!CEC?