Congruence & Similarity
One unit, two big ideas. First, congruent: same shape, same size, and the marks on the figure tell you which parts match. Then similar: same shape, different size, where matching parts are no longer equal but their ratios are. Everything in the second half — scale factor, the side-splitter theorem, the altitude to a hypotenuse — is one idea applied in a new place.
Lesson Map
This is the order the unit is taught in class. Each row names the one skill that shows up on the assessment.
| Lesson | Focus | The skill being checked |
|---|---|---|
| Lesson 1 | Congruent figures and corresponding parts | Read a congruence statement in order and name the matching sides and angles. |
| Lesson 2 — Day 1 | Classifying triangles; the Isosceles Triangle Theorem | Name a triangle by sides and by angles, and use base angles \(\cong\) legs in both directions. |
| Lesson 2 — Day 2 | Medians and altitudes | Tell a median from an altitude, and know when one segment is both. |
| Lesson 3 | Similar figures and scale factor | Find the scale factor from one matching pair, then use it on every other pair. |
| Lesson 4 | Parallel lines inside triangles | Apply the Triangle Proportionality Theorem, and use its converse to test for parallel. |
| Lesson 5 — Day 1 | Similarity in right triangles | Split a right triangle with the altitude to the hypotenuse and name the three similar triangles. |
| Lesson 5 — Day 2 | Geometric mean and applications | Use \(h=\sqrt{pq}\) and the leg relationships to find a missing length. |
| Inquiry | No Ladder Required | Measure a height you cannot reach five independent ways, then defend one number. |
| Assessment | Unit 2 review sheet | Twelve questions in assessment order, with an answer key and a reteach guide. |
Content Summary
The idea first, then the method, then a worked example in the format you are expected to reproduce.
Congruent Figures and Corresponding Parts
Two figures are congruent when one can be moved onto the other by rigid motions alone — the translations and reflections from Unit 1, plus rotations. Nothing stretches. Every side keeps its length and every angle keeps its measure.
That gives the working definition used all year: two figures are congruent exactly when all corresponding sides are congruent and all corresponding angles are congruent. The word doing the work is corresponding, and the congruence statement is what tells you which parts correspond.
Read it position by position. First letter with first letter, second with second, third with third. From that pairing everything else follows: \(\overline{AB}\cong\overline{XY}\), \(\overline{BC}\cong\overline{YZ}\), \(\overline{CA}\cong\overline{ZX}\), \(\angle A\cong\angle X\), \(\angle B\cong\angle Y\), \(\angle C\cong\angle Z\).
One tick matches one tick, one arc matches one arc. The marks — not the picture — tell you which vertex pairs with which, so this is \(\triangle ABC\cong\triangle XZY\).
The order matters and it is not negotiable. \(\triangle ABC\cong\triangle XYZ\) and \(\triangle ABC\cong\triangle XZY\) say two different things, and at most one of them is true. When you write a congruence statement yourself, find the matching vertices first and then write the letters in that order.
Congruence is not only for triangles. \(ABCD\cong KGNP\) says \(A\) pairs with \(K\), \(B\) with \(G\), \(C\) with \(N\), and \(D\) with \(P\).
Classifying Triangles
Every triangle gets two names: one from its angles, one from its sides. You need both.
Every triangle gets exactly one of these four angle names.
Every equilateral triangle is also isosceles, because “isosceles” asks for at least two congruent sides.
Two facts about these names cause most of the errors:
- A triangle has at most one right or obtuse angle, because the three angles must total \(180^\circ\).
- Equilateral is a special case of isosceles, not an alternative to it. “Isosceles” means at least two congruent sides.
The Isosceles Triangle Theorem
An isosceles triangle has its own vocabulary, and the theorem is stated in those words.
The two congruent sides are the legs. The third side is the base, and the two angles that touch it are the base angles.
Read left to right, that is the Isosceles Triangle Theorem: congruent legs force congruent base angles. Read right to left, that is the converse: congruent base angles force congruent legs. Both directions are used on the assessment, and problems are written specifically to check that you can go the second way.
Worked example. In \(\triangle ABC\), \(\overline{AB}\cong\overline{AC}\) and the vertex angle \(\angle A=48^\circ\). Find the base angles.
The vertex angle is subtracted from \(180\) first, and only then split in half. Splitting \(180\) in half first is the most common wrong answer on this question.
One more consequence, used constantly in the second half of the unit: in an isosceles triangle, the segment drawn from the vertex angle to the base does three things at once.
One segment, three jobs: it bisects the vertex angle, it bisects the base, and it is perpendicular to the base.
An equilateral triangle is isosceles three different ways, so all three angles are congruent, and each measures \(60^\circ\).
Medians and Altitudes
Both are segments drawn from a vertex to the opposite side. They are told apart by what happens at the other end.
Median
From a vertex to the midpoint of the opposite side. Defined by equal lengths — look for tick marks on the side it lands on.
Altitude
From a vertex perpendicular to the line containing the opposite side. Defined by a right angle — look for the square corner.
A median hits the midpoint. An altitude hits at a right angle. In an isosceles triangle the segment from the vertex angle is both at once — in every other triangle they are different segments.
Every triangle has three of each. The three medians meet at the centroid, which is the balance point of the triangle; the three altitudes meet at the orthocenter. A segment can be both at once, but only when the triangle is isosceles and the segment comes from the vertex angle — which is exactly the three-jobs picture from Day 1.
In an obtuse triangle two of the three altitudes land outside the triangle, on an extension of the side. That is still an altitude.
Similar Figures and Scale Factor
Congruent figures are the same shape and the same size. Similar figures are the same shape and any size. The definition changes in exactly one place:
Congruent \((\cong)\)
Corresponding angles are congruent.
Corresponding sides are congruent.
Similar \((\sim)\)
Corresponding angles are congruent.
Corresponding sides are proportional.
The constant ratio between corresponding sides is the scale factor, written \(k\).
Same shape, different size. Every pair of corresponding sides has the same ratio: \(\tfrac{9}{6}=\tfrac{12}{8}=\tfrac{15}{10}=1.5\).
Worked example. \(\triangle ABC\sim\triangle DEF\) with \(AB=6\), \(BC=8\) and \(DE=9\). Find \(EF\).
Use the one pair you know completely to get \(k\), then apply \(k\) everywhere else.
You can also set up a proportion and cross multiply, \(\tfrac{6}{9}=\tfrac{8}{EF}\), which gives the same \(12\). Either method is accepted; finding \(k\) once is faster when a problem asks for several sides.
Two facts about scale factor that get tested directly:
- Perimeter scales by \(k\). If \(k=2.5\), the perimeter is \(2.5\) times as large.
- Area scales by \(k^2\). If \(k=2.5\), the area is \(6.25\) times as large.
Parallel Lines Inside Triangles
Draw a line across a triangle parallel to one side. It creates a small triangle at the top that is similar to the whole triangle, because the parallel line copies the angles down. The similarity first gives proportional corresponding sides, such as \(\tfrac{AD}{AB}=\tfrac{AE}{AC}\). From that relationship, the two sides are also divided proportionally, which produces the Triangle Proportionality Theorem, also called the Side-Splitter Theorem.
\(\overline{DE}\parallel\overline{BC}\), so the two sides it crosses are cut in the same ratio: \(\tfrac{AD}{DB}=\tfrac{AE}{EC}\).
Worked example. \(\overline{DE}\parallel\overline{BC}\), \(AD=8\), \(DB=12\), \(AE=10\). Find \(EC\).
The converse is just as important, and it is the only tool you have for proving two segments are parallel in this unit:
To use it, compute both ratios as decimals and compare. Equal means parallel. Not equal means not parallel — and that is a complete, correct answer.
Similarity in Right Triangles
Take a right triangle and drop the altitude from the right angle to the hypotenuse. Something unusual happens: the two triangles it creates are similar to each other, and both are similar to the triangle you started with. Three similar triangles from one line.
The altitude \(\overline{CD}\) cuts the hypotenuse into \(p\) and \(q\), and cuts the triangle into two smaller triangles that are similar to each other and to the original.
Redrawn so the matching parts line up: short leg on the left, long leg on the bottom, hypotenuse across the top. Reading down a column gives every proportion in this lesson.
Rather than rebuilding the proportions each time, memorize the three relationships they produce. \(p\) and \(q\) are the two pieces of the hypotenuse, \(h\) is the altitude, and each leg sits next to its own piece.
The altitude
The altitude is the geometric mean of the two pieces of the hypotenuse.
Each leg
A leg is the geometric mean of the whole hypotenuse and the piece next to that leg.
Geometric mean
“Geometric mean” just names the middle term of a proportion whose two outer terms are \(p\) and \(q\).
Worked example. The altitude to the hypotenuse cuts it into pieces of \(4\) and \(9\). Find the altitude, and find the leg next to the piece of length \(4\).
Check it with Unit 1: the other leg is \(\sqrt{9\cdot 13}=\sqrt{117}\approx 10.82\), and \(52+117=169=13^2\). The Pythagorean Theorem and these relationships always agree — in fact one standard proof of the Pythagorean Theorem is built out of exactly this figure.
Vocabulary
These are the words used in the questions. Misreading one of them is the same as not knowing the math.
Same shape and same size. All corresponding sides and all corresponding angles are congruent.
The sides and angles that are matched to each other by a congruence or similarity statement. The letter order tells you the matching.
A statement such as \(\triangle ABC\cong\triangle XYZ\). Position by position, it names which vertex pairs with which.
A translation, reflection, or rotation. Two figures are congruent exactly when rigid motions carry one onto the other.
A triangle with no congruent sides.
A triangle with at least two congruent sides.
A triangle with three congruent sides. Also isosceles, and always equiangular with three \(60^\circ\) angles.
Named by its largest angle: all angles under \(90^\circ\), exactly one \(90^\circ\) angle, or one angle over \(90^\circ\).
The two congruent sides are the legs; the third side is the base.
The two angles that touch the base of an isosceles triangle. The Isosceles Triangle Theorem says they are congruent.
The angle between the two legs of an isosceles triangle — the one that is not a base angle.
A segment from a vertex to the midpoint of the opposite side.
A segment from a vertex perpendicular to the line containing the opposite side. In an obtuse triangle it may land outside the triangle.
The point where the three medians meet. The balance point of the triangle.
Same shape, any size. Corresponding angles are congruent and corresponding sides are proportional.
The constant ratio between corresponding sides of similar figures. Perimeter scales by \(k\); area scales by \(k^2\).
An equation stating two ratios are equal, such as \(\tfrac{a}{b}=\tfrac{c}{d}\). Solve it by cross multiplying.
A line parallel to one side of a triangle divides the other two sides proportionally.
If a line divides two sides of a triangle proportionally, then it is parallel to the third side.
The value \(h\) with \(\tfrac{p}{h}=\tfrac{h}{q}\), so \(h=\sqrt{pq}\). The middle term of a proportion.
Finding a length you cannot reach by measuring lengths you can reach and using similar triangles.
Sample Problems
Eighteen problems in homework and assessment format, grouped by lesson. Work each one on paper first — the solutions are one click away, which makes them very easy to read too early. Figures are not drawn to scale.
Lesson 1 — Congruent Figures
Solve for the variables.
Given \(\triangle ABC\cong\triangle XYZ\), find \(x\), \(y\), and \(z\).
Show solution
Match by letter position: \(\overline{AB}\) with \(\overline{XY}\), \(\overline{BC}\) with \(\overline{YZ}\), \(\overline{CA}\) with \(\overline{ZX}\). Corresponding sides of congruent triangles are equal, so set each pair equal and solve.
Check: the sides of \(\triangle ABC\) are \(11\), \(11\), \(12\) — the same three lengths as \(\triangle XYZ\).
Find the missing angle measures.
Given \(\triangle DEF\cong\triangle RST\), find \(x\), \(y\), and \(z\).
Show solution
The third angle of \(\triangle DEF\) comes from the angle sum, and then corresponding angles are copied straight across.
Now match by position: \(D\leftrightarrow R\), \(E\leftrightarrow S\), \(F\leftrightarrow T\).
Check: \(91+57+32=180\).
Lesson 2 Day 1 — Classification and Isosceles Triangles
Find both missing angles.
In \(\triangle ABC\), \(\overline{AB}\cong\overline{AC}\) and \(m\angle B=42^\circ\). Find \(m\angle C\) and \(m\angle A\).
Show solution
The tick marks put the congruent sides at \(\overline{AB}\) and \(\overline{AC}\), so the base is \(\overline{BC}\) and the base angles are \(\angle B\) and \(\angle C\). By the Isosceles Triangle Theorem they are congruent.
The vertex angle is whatever is left over.
The triangle is isosceles and obtuse.
Solve for \(x\), then find every angle.
\(\triangle RST\) has \(\overline{RS}\cong\overline{RT}\). Find \(x\), then find all three angle measures.
Show solution
Both expressions sit on base angles, so the Isosceles Triangle Theorem makes them equal.
Substitute back into both expressions — if they do not agree, the algebra is wrong.
Use the vertex angle bisector.
In isosceles \(\triangle ABC\), \(\overline{AD}\) bisects the vertex angle \(\angle A\). Find \(x\) and the length of the base \(\overline{BC}\).
Show solution
In an isosceles triangle the bisector of the vertex angle also bisects the base, so \(D\) is the midpoint of \(\overline{BC}\) and the two halves are equal.
The question asks for the whole base, not one half. Stopping at \(5\) is the most common error here.
Use the definition of a median.
\(\overline{SZ}\) is a median of \(\triangle STU\). Find \(x\) and the length of \(\overline{TU}\).
Show solution
A median goes to the midpoint, so \(TZ=ZU\). That single fact is the whole equation.
Lesson 2 Day 2 — Medians and Altitudes
Median, altitude, both, or neither?
In each triangle, \(\overline{AD}\) is drawn from vertex \(A\) to side \(\overline{BC}\). Classify each segment as a median, an altitude, both, or neither.
Show solution
Read only the marks. Tick marks on \(\overline{BD}\) and \(\overline{DC}\) mean midpoint, so median. A square corner at \(D\) means perpendicular, so altitude.
- a. Tick marks, no right angle → median.
- b. Tick marks and a right angle → both. This is the isosceles case.
- c. Right angle, no tick marks → altitude.
- d. No tick marks, no right angle → neither.
“Neither” is a legitimate answer. A segment from a vertex to the opposite side is only special if it is marked as special.
Is it an altitude?
Segment \(\overline{AD}\) is drawn from \(A\) to \(\overline{BC}\). Using only the angles given, decide whether \(\overline{AD}\) is an altitude of \(\triangle ABC\). Justify the answer.
Show solution
\(\overline{AD}\) is an altitude exactly when \(\angle ADB\) measures \(90^\circ\). Find that angle from \(\triangle ABD\), which has all the information needed.
\(91^\circ\neq 90^\circ\), so \(\overline{AD}\) is not an altitude. It misses perpendicular by one degree.
This is the reason the figure cannot be trusted by eye. One degree is invisible on paper and decisive in the answer.
Lesson 3 — Similar Figures and Scale Factor
Write the similarity statement.
The angle arcs show which angles are congruent. Explain why the triangles are similar, and write the similarity statement.
Show solution
Matching arcs mark congruent angles: \(\angle A\cong\angle P\) (one arc), \(\angle B\cong\angle Q\) (two arcs), and therefore \(\angle C\cong\angle R\), since the third angles are what is left of \(180^\circ\) in each triangle.
All three pairs of corresponding angles are congruent, which is the definition of similar figures for triangles.
Note that no side lengths were needed. For triangles, congruent angles are enough — the proportional sides come along automatically.
Find the scale factor, then the missing side.
Quadrilateral \(CARS\sim\) quadrilateral \(DOTE\). Find the scale factor from \(CARS\) to \(DOTE\), then find \(OT\).
Show solution
Letter order gives the correspondence: \(C\leftrightarrow D\), \(A\leftrightarrow O\), \(R\leftrightarrow T\), \(S\leftrightarrow E\). So \(\overline{CA}\) matches \(\overline{DO}\), and \(\overline{AR}\) matches \(\overline{OT}\).
The scale factor is a single number that works on every pair of corresponding sides. Find it once and reuse it.
Find two missing sides.
\(\triangle ABC\sim\triangle XYZ\). Find \(YZ\) and \(ZX\).
Show solution
The one complete pair is \(\overline{AB}\) with \(\overline{XY}\).
Check with Unit 1: \(9\)–\(12\)–\(15\) is a Pythagorean triple, the \(3\)–\(4\)–\(5\) family again. A right triangle scaled up stays a right triangle.
Lesson 4 — Parallel Lines Inside Triangles
Apply the Side-Splitter Theorem.
In \(\triangle ABC\), \(\overline{DE}\parallel\overline{BC}\). Find \(x\).
Show solution
\(\overline{DE}\) is parallel to \(\overline{BC}\), so it splits the two sides it crosses in the same ratio. Use piece over piece on both sides.
Check: \(\tfrac{6}{9}=0.667\) and \(\tfrac{8}{12}=0.667\).
Is the segment parallel?
In \(\triangle PQR\), point \(S\) is on \(\overline{PQ}\) and point \(T\) is on \(\overline{PR}\). Determine whether \(\overline{ST}\parallel\overline{QR}\). Show the work that justifies the answer.
Show solution
This is the converse. Compute both ratios and compare — do not solve anything.
The ratios are equal, so by the converse of the Triangle Proportionality Theorem, \(\overline{ST}\parallel\overline{QR}\).
Both fractions reduce to \(\tfrac{2}{3}\), which is a cleaner way to show it than decimals if the numbers cooperate.
Find the whole side.
In \(\triangle ABC\), \(\overline{DE}\parallel\overline{BC}\). Find \(x\), the length of the entire side \(\overline{AC}\).
Show solution
The unknown is a whole side, so use the piece-over-whole form on both sides. First build the whole side you do know: \(AB=4+8=12\).
Check with piece over piece: \(EC=15-5=10\), and \(\tfrac{4}{8}=\tfrac{5}{10}\). Both forms agree, as long as you do not mix them inside one proportion.
Lesson 5 — Similarity in Right Triangles
Find the altitude.
\(\overline{CD}\) is the altitude to the hypotenuse of right \(\triangle ABC\). Find \(x\).
Show solution
The altitude to the hypotenuse is the geometric mean of the two pieces it creates.
Equivalently, from the proportion \(\tfrac{4}{x}=\tfrac{x}{9}\), cross multiplying gives \(x^2=36\).
Find both legs.
The hypotenuse of right \(\triangle ABC\) measures \(20\), and the altitude from \(C\) meets it at \(D\) with \(AD=5\). Find \(x=AC\) and \(y=CB\). Round to two decimal places where needed.
Show solution
First find the other piece of the hypotenuse: \(DB=20-5=15\). Each leg is the geometric mean of the whole hypotenuse and the piece next to that leg.
Check with the Pythagorean Theorem: \(10^2+300=100+300=400=20^2\).
Find all three.
The altitude to the hypotenuse divides it into pieces of \(9\) and \(16\). Find \(x\), \(y\), and \(z\).
Show solution
Write the whole hypotenuse on the figure first: \(AB=9+16=25\).
Check: \(15\)–\(20\)–\(25\) is the \(3\)–\(4\)–\(5\) triple times five. Every one of these problems has a Unit 1 check available — use it.
Application: the roof truss.
A roof truss has a \(24\) ft horizontal tie beam from \(L\) to \(R\). The two rafters meet at the peak \(P\) in a right angle, and a vertical support runs from the peak straight down to the tie beam at \(F\), which is \(8\) ft from \(L\). How long is the vertical support? Round to two decimal places.
Show solution
Translate the situation before touching the algebra. The right angle is at the peak, the tie beam is the hypotenuse, and the vertical support is the altitude drawn to it. That makes this problem 15 in a costume.
The support is about \(11.31\) feet, or roughly \(11\) ft \(4\) in.
Equations & Rules
Everything the unit needs on one page. Print it and keep it in the notebook — none of these relationships are provided on the assessment.
Congruence
Definition
Corresponding sides congruent and corresponding angles congruent. Rigid motions carry one figure onto the other.
Reading a statement
First with first, second with second, third with third. The order is the information.
Angle sum
Used in almost every problem to recover the third angle.
Triangle classification
| By angles | Condition | By sides | Condition |
|---|---|---|---|
| Acute | All three angles under \(90^\circ\) | Scalene | No congruent sides |
| Right | Exactly one \(90^\circ\) angle | Isosceles | At least two congruent sides |
| Obtuse | One angle over \(90^\circ\) | Equilateral | Three congruent sides |
| Equiangular | Three \(60^\circ\) angles | Equilateral and equiangular describe the same triangle. | |
Isosceles triangles, medians, and altitudes
Isosceles Triangle Theorem
Congruent legs give congruent base angles.
Converse
Congruent base angles give congruent legs. Both directions are tested.
Base angles from the vertex angle
Subtract first, then halve. Never halve \(180\).
Median
Vertex to the midpoint of the opposite side. Three medians meet at the centroid.
Altitude
Vertex perpendicular to the opposite side. Three altitudes meet at the orthocenter.
Both at once
The same segment bisects the vertex angle, bisects the base, and is perpendicular to it.
Similarity
Definition
The angles behave exactly as in congruence; only the sides change.
Scale factor
One number, used on every corresponding pair.
Perimeter and area
Perimeter scales by \(k\); area scales by \(k\) squared.
Proportions in triangles
| Situation | Relationship | Use it when |
|---|---|---|
| Side-splitter, piece over piece | \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\) | Both pieces of a side are labeled. |
| Side-splitter, piece over whole | \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\) | The unknown is an entire side. |
| Converse of the side-splitter | ratios equal \(\Rightarrow\) parallel | The question asks whether two segments are parallel. |
| Altitude to the hypotenuse | \(h=\sqrt{pq}\) | You want the altitude from the two pieces. |
| Leg of a right triangle | \(a=\sqrt{p\cdot c}\) | You want a leg. \(p\) touches that leg; \(c\) is the whole hypotenuse. |
| Three similar triangles | \(\triangle ACB\sim\triangle ADC\sim\triangle CDB\) | You are asked to name them or to justify a proportion. |
Congruent and similar: what is shared
- Corresponding angles are congruent in both.
- Both depend on a correspondence, given by letter order.
- Both are written as statements, not pictures.
Congruent and similar: what differs
- Congruent: corresponding sides are equal, so \(k=1\).
- Similar: corresponding sides are proportional, and \(k\) can be anything positive.
- Congruence preserves area; similarity multiplies it by \(k^2\).
Before you turn in the assessment
- Every congruence and similarity statement written with the letters in matching order.
- Every variable substituted back into both expressions it came from.
- Angles in each triangle adding to \(180^\circ\).
- Whole side or piece of a side — checked before writing each proportion.
- The whole hypotenuse written on the figure before any leg is computed.
- Every radical either simplified exactly or rounded to two decimal places.
- Parallel questions answered with two computed ratios and a sentence, not a guess.
Video Library
Walkthroughs matched to the lessons in this unit, mostly Khan Academy. Use them after a first attempt, not instead of one. Videos do not appear on the printed copy.
Lesson 1 — Congruent Figures
Lesson 2 Day 1 — Classification and Isosceles Triangles
Lesson 2 Day 2 — Medians and Altitudes
Lesson 3 — Similar Figures and Scale Factor
Lesson 4 — Parallel Lines Inside Triangles
Lesson 5 — Similarity in Right Triangles
Applications — Measuring What You Cannot Reach
These four go with the inquiry project. Watch at least one before the build day.
Where to Go Next
The other two Unit 2 pages use the same skills in different formats.
Unit 2 Review Sheet
Twelve assessment-style questions with figures, worked solutions, an answer key, and a reteach guide that points back to the lesson for anything missed.
Open the review sheet →Inquiry Project: No Ladder Required
Build a \(45^\circ\) sighting wedge, then measure something too tall to reach five independent ways — a 45° sight, shadows, a mirror, a sighting stick, and a geometric mean — and defend one final number.
Open the project →