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Triangle \(ABC\) is congruent to triangle \(EDF\). So, Kiran knows that there is a sequence of rigid motions that takes \(ABC\) to \(EDF\).
Select all true statements after the transformations:
Angle \(A\) coincides with angle \(F\).
Angle \(B\) coincides with angle \(D\).
Segment \(AC\) coincides with segment \(EF\).
Segment \(BC\) coincides with segment \(ED\).
Segment \(AB\) coincides with segment \(ED\).
A rotation by angle \(ACE\) using point \(C\) as the center takes triangle \(CBA\) onto triangle \(CDE\).
The triangles are congruent. Which sequence of rigid motions will take triangle \(XYZ\) onto triangle \(BCA\)?
Translate \(XYZ\) using directed line segment \(YC\). Rotate \(X’Y’Z’\) using \(C\) as the center so that \(X’\) coincides with \(B\). Reflect \(X’’Y’’Z’’\) across line \(CB\).
Translate \(XYZ\) using directed line segment \(YC\). Rotate \(X’Y’Z’\) using \(C\) as the center so that \(X’\) coincides with \(B\). Reflect \(X’’Y’’Z’’\) across line \(AC\).
Translate \(XYZ\) using directed line segment \(YC\). Rotate \(X’Y’Z’\) using \(C\) as the center so that \(X’\) coincides with \(A\). Reflect \(X’’Y’’Z’’\) across line \(CB\).
Translate \(XYZ\) using directed line segment \(YC\). Rotate \(X’Y’Z’\) using \(C\) as the center so that \(X’\) coincides with \(A\). Reflect \(X’’Y’’Z’’\) across line \(AC\).
Triangle \(HEF\) is the image of triangle \(FGH\) after a 180-degree rotation around point \(K\). Select all statements that must be true.
Triangle \(HGF\) is congruent to triangle \(FEH\).
Triangle \(GFH \) is congruent to triangle \(EFH\).
Angle \(KHE\) is congruent to angle \(KHG\).
Angle \(GHK\) is congruent to angle \(EFK\).
Segment \(EH\) is congruent to segment \(GH\).
Segment \(HG\) is congruent to segment \(FE\).
Segment \(FH\) is congruent to segment \(HF\).
Line \(SD\) is a line of symmetry for figure \(ASMHZDPX\). Tyler says that \(ASDPX\) is congruent to \(SMDZH\) because sides \(AS\) and \(MS\) are corresponding.
Triangle \(ABC\) is congruent to triangle \(DEF\). Select all the statements that are a result of corresponding parts of congruent triangles being congruent.
Segment \(AC\) is congruent to segment \(EF\).
Segment \(BC\) is congruent to segment \(EF\).
Angle \(BAC\) is congruent to angle \(EDF\).
Angle \(BCA\) is congruent to angle \(EDF\).
Angle \(CBA\) is congruent to angle \(FED\).
When triangle \(ABC\) is reflected across line \(AB\), the image is triangle \(ABD\). Why is angle \(ACD\) congruent to angle \(ADB\)?
Corresponding parts of congruent figures are congruent.
Congruent parts of congruent figures are corresponding.
Segment \(AB\) is a perpendicular bisector of segment \(DC\).
An isosceles triangle has a pair of congruent angles.