Geometry: Rigid Motions & Congruence

How to do it:

1. Write definitions using point, line, and distance, not "looks like."

2. A line segment is the set of points between two endpoints, including them. A ray has one endpoint.

3. Parallel lines never meet. A circle is all points at a fixed distance (radius) from a center.

Worked example: Perpendicular lines meet at a right angle. A right angle is congruent to the angle that sits next to it on a straight line, so each measures $90°\text{.}$

Check: "these lines look parallel on my paper" is not a definition. You need equal corresponding angles from a transversal, or a slope argument later.

How to do it:

1. Name a translation, rotation, or reflection that matches one figure to the other.

2. Rigid motions preserve distance and angle measure, so the figures are congruent.

3. Then corresponding parts are congruent (CPCTC after a proof).

Worked example: Reflect triangle $ABC$ over line $m$ so $A$ maps to $A'\text{.}$ If $B$ and $C$ land on $B'$ and $C'\text{,}$ then $\triangle ABC \cong \triangle A'B'C'\text{.}$

Check: a dilation is not rigid; it changes size. That is similarity, next topic.

How to do it:

1. Mark the parts you know: sides, included angle, or included side.

2. ASA, SAS, and SSS each force a unique triangle up to congruence.

3. Name the criterion and list corresponding vertices in order.

Worked example: Two sides and the included angle (SAS) determine a unique triangle, so another triangle with the same SAS is congruent.

Check: SSA is not a congruence shortcut (ambiguous case). HL works for right triangles; do not use it on acute triangles.

How to do it:

1. Draw the two lines and the transversal. Label the eight angles.

2. If the lines are parallel, corresponding angles are congruent, and so are alternate interior angles.

3. Adjacent angles on a straight line still sum to $180°\text{.}$ Vertical angles are congruent even without parallels.

Worked example: If $m\angle 1 = 110°$ and $\angle 2$ corresponds to $\angle 1\text{,}$ then $m\angle 2 = 110°\text{.}$ The adjacent angle on the straight line is $70°\text{.}$

Check: corresponding angles are equal because the lines are parallel (or you are proving they are). Do not assume parallelism from a sloppy sketch.

How to do it:

1. Translation by $\langle a,b\rangle\text{:}$ $(x,y)\to(x+a,\,y+b)\text{.}$

2. Reflections: over the $x$-axis $(x,y)\to(x,-y)\text{;}$ over the $y$-axis $(x,y)\to(-x,y)\text{;}$ over the line $y=x\text{,}$ $(x,y)\to(y,x)\text{.}$

3. Rotations about the origin: $90°$ counterclockwise $(x,y)\to(-y,x)\text{;}$ $180°$ $(x,y)\to(-x,-y)\text{;}$ $270°$ counterclockwise $(x,y)\to(y,-x)\text{.}$

Worked example: rotate $A(1,2)\text{,}$ $B(4,2)\text{,}$ $C(4,6)$ by $90°$ counterclockwise about the origin: $A'(-2,1)\text{,}$ $B'(-2,4)\text{,}$ $C'(-6,4)\text{.}$ Reflect $B(4,2)$ over the $y$-axis: $B''(-4,2)\text{.}$

Check: a rigid motion keeps every distance. $AB=3$ and $A'B'=3\text{,}$ so the rotation passed the test. A rule like $(x,y)\to(2x,2y)$ doubles distances, so it is a dilation, not a rigid motion.

How to do it:

1. Draw a line through one vertex parallel to the opposite side. Alternate interior angles copy the other two angles onto a straight line, so the three angles add to $180°\text{.}$

2. An exterior angle (extend one side) equals the sum of the two remote interior angles.

3. Use both facts to find missing angles without measuring.

Worked example: a triangle has angles $50°$ and $60°\text{,}$ so the third is $70°\text{.}$ Extend a side at that third vertex: the exterior angle is $180°-70°=110°\text{,}$ which is also $50°+60°\text{.}$

Check: the three interior angles must add to exactly $180°\text{,}$ so no triangle has two right angles or two obtuse angles. Each exterior angle is bigger than either remote interior angle.

How to do it:

1. List the givens and mark them on a diagram. Look for a shared side (reflexive property) or vertical angles.

2. Match a criterion: ASA, SAS, SSS, AAS, or HL for right triangles. Name the triangles with vertices in corresponding order.

3. Once the triangles are congruent, CPCTC (corresponding parts of congruent triangles are congruent) gives any remaining side or angle.

Worked example: in isosceles $\triangle ABC$ with $AB\cong CB\text{,}$ draw the bisector $BD$ of $\angle ABC\text{.}$ Then $AB\cong CB$ (given), $\angle ABD\cong\angle CBD$ (bisector), $BD\cong BD$ (reflexive), so $\triangle ABD\cong\triangle CBD$ by SAS. By CPCTC, $\angle A\cong\angle C\text{:}$ the base angles of an isosceles triangle are congruent.

Check: every statement needs a reason: a given, a definition, a postulate, or a theorem already proved. Writing $\triangle ABD\cong\triangle DBC$ with the vertices out of order breaks CPCTC.

How to do it:

1. In a parallelogram, opposite sides are congruent, opposite angles are congruent, and consecutive angles are supplementary.

2. The diagonals bisect each other. Each diagonal splits the parallelogram into two congruent triangles (ASA, using alternate interior angles).

3. Special cases: a rectangle has congruent diagonals, a rhombus has perpendicular diagonals, and a square has both.

Worked example: parallelogram $ABCD$ has $AB=8\text{,}$ $BC=5\text{,}$ and $\angle A=110°\text{.}$ Then $CD=8\text{,}$ $DA=5\text{,}$ $\angle C=110°\text{,}$ and $\angle B=\angle D=70°\text{.}$ If the diagonals meet at $E$ and $AE=6\text{,}$ then $EC=6\text{.}$

Check: to prove a quadrilateral is a parallelogram, one pair of opposite sides that are both parallel and congruent is enough. Congruent diagonals alone do not make a rectangle unless the shape is already a parallelogram.

How to do it:

1. Copy a segment: set the compass to its length and swing that arc on a ray.

2. Bisect an angle: arc from the vertex across both sides, then equal arcs from those two points; connect the vertex to where they cross.

3. Perpendicular bisector: from each endpoint, swing arcs of the same radius (more than half the segment) above and below; connect the two crossings.

Worked example: to find the center of a circle through three points $A\text{,}$ $B\text{,}$ $C\text{,}$ construct the perpendicular bisectors of $AB$ and $BC\text{;}$ they meet at the center, because every point on a perpendicular bisector is the same distance from both endpoints.

Check: a construction never uses the marks on a ruler or a protractor. Each step is justified by congruent triangles, which is why the angle-bisector construction really halves the angle (SSS).