Similarity and Proportion: Chapter Review

Vocabulary and notation

Preimage and image

The preimage is the original figure; the image is the transformed figure. A prime, as in $P'$, names an image point.

Dilation and positive scale factor

A dilation resizes from one center using one positive factor $k$ in every direction. Enlargement: factor greater than one; reduction: between zero and one; identity: one.

Fixed point

A point whose image is its own location. A line remaining unchanged as a set does not mean every point on it is fixed.

Similarity

A positive dilation together with translations, rotations, or reflections maps one whole figure exactly onto the other. Congruence is included at factor $1$.

Correspondence

The ordered matching of vertices, angles, and side endpoints. In $\triangle ABC\sim\triangle PQR$, the pairs are $A$ with $P$, $B$ with $Q$, and $C$ with $R$.

Proportional lengths

Matched length ratios agree in one consistent direction. A ratio of two sides within one triangle need not be the factor between two triangles.

Included angle

The angle formed by the two selected sides, at their common endpoint. For sides $AB$ and $AC$, it is $\angle BAC$.

Midsegment

A segment joining the midpoints of two triangle sides. The three midsegments enclose the central, half-size triangle.

Internal angle bisector

A segment from a triangle vertex to the opposite side that divides the interior angle into two equal angles. Its endpoint lies strictly inside that side.

Uniform map scale

One length ratio applies in every direction. A scale $1:n$ compares drawing length with real length in common units.

Formulas and theorems

• Accepted dilation properties

  $$\begin{gathered} OP'=k\,OP\\ A'B'=k\,AB\\ m\angle A'B'C'=m\angle ABC \end{gathered}$$

  Text description

  The solid blue triangle has the shared center O and vertices A and B. The dashed red image has vertices A prime and B prime twice as far along the same rays. The segment joining the image vertices is also twice as long. The center remains in place.

  Use when One center $O$ and positive factor $k$ apply throughout. For $P\ne O$, the image lies on ray $OP$; $O$ maps to itself. Segments have distinct endpoints and angle rays are defined. Length, angle, and line-image properties are accepted foundations.

• Coordinate dilation about any center

  $$\begin{gathered} x'=a+k(x-a)\\ y'=b+k(y-b) \end{gathered}$$

  Use when The center is $O(a,b)$ and the factor is positive. For the origin, both center coordinates are zero, giving $(x,y)\mapsto(kx,ky)$.

  e.g. For center $(-1,3)$ and factor $\frac32$, find the image of $(5,-1)$: $(8,-3)$.

• Fixed points and line images

  • Identity factor: at $k=1$, every point is fixed and every line maps onto itself.
  • Other positive factors: at $k\ne1$, only $O$ is fixed; a through-center line maps onto itself as a set, while an away-center line maps onto a distinct parallel.

  Text description

  The center O lies on the bottom line, named m and m prime because it maps onto itself as a whole line. The blue line ell lies above O. Its red dashed image lies twice as far above O and is a distinct parallel. The illustrated positive factor is two.

  Use when These statements concern a dilation of the whole plane with center $O$ and positive factor $k$.

• Corresponding sides and reverse factors

  $$\begin{gathered} \frac{PQ}{AB}=k\\ \frac{QR}{BC}=k\\ \frac{PR}{AC}=k\\ k_{\rm reverse}=\frac1k \end{gathered}$$

  Text description

  The blue triangle has A at its lower left corner, B to the right, and C above A. The larger red dashed triangle has P at its upper right corner, Q to the left, and R below P. The matching is A to P, B to Q, and C to R despite their different orientations.

  Use when Similarity $\triangle ABC\sim\triangle PQR$ is already established. The positive factor maps the first triangle to the second; corresponding angles are equal. Factor one means congruence.

• AA similarity

  Two equal corresponding angle pairs at distinct vertices establish triangle similarity.

  Use when Both triangles are nondegenerate. The angle sum supplies a third angle match; proportional sides follow from similarity, not from the angle sum alone.

• SSS similarity

  Three corresponding side ratios equal to one positive factor establish triangle similarity.

  Use when Both side triples must form nondegenerate triangles. Preserve one vertex matching and the same ratio direction for all three pairs.

  e.g. Test side triples $7,11,15$ and $9.8,15.4,21$: similar by SSS, with first-to-second factor $1.4$.

• SAS similarity

  Two equal positive corresponding side ratios and equal included angles establish triangle similarity.

  Use when Both triangles are nondegenerate. The equal angles must lie between the two sides actually compared in each triangle.

• Side-splitter theorem and converse

  Equal part-to-part ratios hold exactly for a cut parallel to the base. Such a cut also gives the whole-side equalities.

  $$\begin{gathered} \frac{AD}{DB}=\frac{AE}{EC}\\ \frac{AD}{AB}=\frac{AE}{AC}\\ \frac{DE}{BC}=\frac{AD}{AB} \end{gathered}$$

  Text description

  Triangle ABC has apex A. Point D is strictly between A and B, and point E is strictly between A and C. Segment DE is parallel to BC, as indicated by matching arrow marks. The smaller triangle ADE is dashed red and the whole triangle ABC is solid blue. DB and EC are the leftover side parts.

  Use when Triangle $ABC$ is nondegenerate, $D$ lies strictly between $A,B$, and $E$ strictly between $A,C$. The small triangle is $ADE$; $DB$ and $EC$ are leftover parts.

  e.g. In an interior parallel cut, $AD,DB,AE$ are $5,8,7.5$ cm; find $EC$: $12$ cm.

• Midsegment theorem and central correspondence

  $$\begin{gathered} DE\parallel BC\\ DE=\frac12BC\\ \triangle DEF\sim\triangle CBA \end{gathered}$$

  Text description

  The blue outer triangle ABC is scalene. D is the midpoint of AB, E is the midpoint of AC, and F is the midpoint of BC. One, two, and three tick families respectively mark the equal halves of AB, AC, and BC. The red dashed central triangle has DE parallel to CB, EF parallel to BA, and FD parallel to AC. Its vertices D, E, F match C, B, A.

  Use when In nondegenerate $ABC$, points $D,E,F$ are the midpoints of $AB,AC,BC$, respectively. Each midsegment is parallel to its matching outer side and half as long. Outer-to-central factor: $1/2$.

• Parallel cross-lines and intercepted intervals

  $$\frac{AB}{BC}=\frac{PQ}{QR}$$

  Text description

  Three horizontal parallel cross-lines meet the blue transversal at A, B, and C in order and the red dashed transversal at P, Q, and R in the corresponding order. The gap from A to B matches the gap from P to Q. The next gap from B to C matches the gap from Q to R. Matching arrow marks identify the three cross-lines as parallel.

  Use when Three distinct parallel lines cut two distinct transversals at $A,B,C$ and $P,Q,R$ in corresponding cross-line order. Intervals are positive. If the transversals are parallel, corresponding intervals are equal.

• Internal angle-bisector ratio and converse

  The opposite-side parts have the adjacent-side ratio exactly when the segment is the internal angle bisector.

  $$\begin{gathered} \frac{BD}{DC}=\frac{AB}{AC}\\ BD=\frac{AB}{AB+AC}\,BC\\ DC=\frac{AC}{AB+AC}\,BC \end{gathered}$$

  Text description

  Segment AD runs from vertex A to an interior point D on side BC. Matching arcs mark the two equal angle halves at A. The side AB is shorter than AC, and the neighboring part BD is shorter than DC. The diagram distinguishes equal angle halves from equal opposite-side parts.

  Use when Triangle $ABC$ is nondegenerate and $D$ lies strictly between $B,C$. Match the part beside $B$ with $AB$, and the part beside $C$ with $AC$.

  e.g. For an internal bisector with $AB,AC,BC$ of $13,22,28$ cm, find $BD$: $10.4$ cm.

• Shadow height model

  $$\frac{H}{s}=\frac{h}{t}$$

  Text description

  The small blue triangle has vertical calibration height h and its own ground shadow t. The larger red dashed triangle has vertical object height H and its own shadow s. Both bases are level and both base angles are right angles. Their slanted sunrays are parallel, so the shadow-tip angles match. Matching arrow marks identify the two sunrays as parallel.

  Use when Vertical objects stand on the same level ground; measurements are simultaneous and sunrays are modeled as parallel. Heights $H,h$ match their own base-to-tip shadows $s,t$, respectively; shadows are positive.

• Horizontal mirror model

  $$\frac{TB}{EF}=\frac{MB}{FM}$$

  Text description

  The blue observer triangle has vertical eye segment EF and ground leg FM. The red dashed object triangle has vertical height TB and ground leg BM. The horizontal mirror M lies between F and B. The sight path goes from T through M to E in one vertical plane. Equal marked mirror angles are supplied by the reflection assumption.

  Use when Horizontal mirror $M$ lies strictly between observer foot $F$ and object base $B$ on level ground. Eye $E$ and top $T$ are vertically above them. A common vertical sight plane and physical equal-angle reflection give AA.

• Crossed survey model

  $$\begin{gathered} \triangle TAB\sim\triangle DCB\\ \frac{AT}{CD}=\frac{AB}{BC} \end{gathered}$$

  Text description

  The straight bank contains A, B, and C in that order. The sight line contains target T, point B, and accessible point D in that order. AT and CD are perpendicular to the bank. The blue triangle TAB matches the red dashed triangle DCB: T matches D, A matches C, and B matches itself. AT is the inaccessible perpendicular distance.

  Use when Bank points occur in order $A,B,C$ and sight-line points in order $T,B,D$. Both $AT$ and accessible $CD$ are perpendicular to bank $AC$. Right angles and vertical angles justify AA.

• Map lengths and uniform reprinting

  $$\begin{gathered} L_{\rm real}=nL_{\rm drawing}\\ L_{\rm drawing}=\frac{L_{\rm real}}n\\ n_{\rm new}=\frac nr \end{gathered}$$

  Use when The original uniform scale is $1:n$, with positive $n$ and common length units. A reprint multiplies every drawing length by the same positive factor $r$; its scale is $1:n_{\rm new}$.

• Perimeter scaling

  $$P_{\rm image}=kP_{\rm original}$$

  Use when The polygons are similar with positive length scale factor $k$ from the original to the image and common units. Perimeter adds every matching boundary side once, so it uses the length factor.

Problem types, step by step

Exam traps

• Trap Treating a stretch as a dilation

  Fix One center, one positive factor, and the ray condition must work throughout. Different directional length factors rule out a dilation, even if one angle stays unchanged.

• Trap Multiplying angle measures by the length factor

  Fix Dilations preserve angle measures. Only lengths and perimeter use the positive length factor.

• Trap Using position instead of named correspondence

  Fix Follow the angle and endpoint matches through turns and reflections. Reverse both ratios together, or take the reciprocal of the mapping factor.

• Trap Replacing a similarity justification with the triangle-angle sum

  Fix The angle sum supplies only the third angle. Scale a side to match, then use ASA for the included side or AAS for another side; SSS and SAS use their corresponding earlier congruence criteria.

• Trap Calling insufficient evidence proof of dissimilarity

  Fix One angle, two side ratios alone, or a nonincluded angle do not guarantee similarity. A failed named matching rules out that matching; consider other matchings before ruling out every similarity.

• Trap Mixing a part with its whole

  Fix Name the denominator: $DB$ differs from $AB$, $DC$ from $BC$, and mirror ground leg $MB$ from whole $FB$ or slanted $EM$.

• Trap Assuming an angle bisector halves the opposite side

  Fix For an internal bisector, equal side parts occur exactly when the adjacent sides are equal. The two small triangles are not automatically similar; perpendicular-distance statements are a different topic.

• Trap Measuring instead of constructing exact divisions

  Fix Keep a fixed compass opening for equal steps and construct the parallels. Supplied segment ratios copy those given lengths; they do not assert that every real length is constructible.

• Trap Trusting a model or a distorted print without its hypotheses

  Fix Verify the angle relationships and a common directional scale. Without a needed assumption, the numerical result is not guaranteed; it need not be impossible.

• Trap Reporting a rounded measurement as an exact equality

  Fix Keep precision during calculation; report the final measurement with approximately or $\approx$. Agreement between methods supports a check without proving every modeling assumption.