GEO01: Angle chasing and your first proofs

Prerequisites and observable goals#

Level: Geometry, Module 1 of 28. About 10 hours over two weeks, in five sessions. You need paper, a pencil, a ruler, and a protractor. The protractor is for your guesses only, never for a proof.

0. The diagnostic (about 35 minutes, closed book)#

This module opens by finding out what you already know, so that nothing here wastes your time and nothing is skipped that you need. Work the twelve questions on paper, alone, with no notes, no calculator, and no AI. Write “not sure” rather than guess wildly: an honest blank tells you more than a lucky answer. Time yourself, then mark your answers against the key that opens this chapter’s complete solutions, and write down your score in each group. The routing table after the questions says how to use this module, and what each group’s score means for the modules after it.

This is the track’s first module, so there is nothing to review yet: the diagnostic doubles as this module’s spaced review. From GEO02 on, each module opens with two or three questions drawn from your error log.

Scoring and routing#

Each question scores 1 (D06 only if all five are right; D11 only if every step has a reason). Add the scores in each group.

Fast lane, if A $=3$, B $=2$, and F $\ge1$: read sections 2.1 and 2.8–2.11 in full; skim 2.2–2.7 through their hero figures and toolbox cards, but do every proof and every “Say it back” in them; do drill GEO01.06 only, then all the problems, all the challenges, and the gate. Full lane, otherwise: everything, in order. Whatever your lane, the mastery gate in section 6 is the same.

1. Quick summary#

Five ideas carry this module.

1. A figure is evidence, not proof. Every claim needs a reason that comes from the givens, a definition, a postulate, or a theorem already proved. What a figure seems to show (a midpoint, a right angle, a pair of parallel lines) counts for nothing unless the words say it.
2. Angles at a point add up. Angles along a straight line add to $180^\circ$, angles all the way around a point add to $360^\circ$, and so vertical angles are equal.
3. Parallel lines carry angles. When a line crosses two parallel lines, it makes the same angles at both crossings. This one assumption, the parallel postulate, is what makes plane geometry flat, and every result below rests on it.
4. A triangle’s angles add to $180^\circ$, and an exterior angle equals the two interior angles away from it. These are the most-used moves in all of angle chasing.
5. Angle chasing is bookkeeping. Name one unknown, pass facts through the figure one mark at a time with a reason for each, close the loop with an equation, then write it up as a chain of “because” steps.

What you will be able to do. By the end you can find any angle in a figure built from lines, triangles, and polygons, and prove your answer in a write-up a grader would accept. The Angle Explorer, on the companion website at bezhig.com/book/explorers/geo01-angles.html, lets you drag the corners of triangles, parallel-line figures, and stars and watch the angle sums stay fixed while every individual angle changes. Use it after section 2.4, and again before the challenges.

2. Deep explanation#

2.1. The rules of the game: what a proof may use#

Geometry has a small vocabulary. Learn each word once, exactly, because proofs lean on the exact meaning.

• A point marks a location and has no size. A line is perfectly straight and runs forever in both directions. A segment $AB$ is the part of a line between the points $A$ and $B$, which are its endpoints; its length is written $AB$ too. A ray $AB$ starts at $A$, passes through $B$, and runs on forever.
• An angle is two rays with the same starting point, called its vertex. The angle between rays $BA$ and $BC$ is written $\angle ABC$. The middle letter is always the vertex: $\angle ABC$ sits at $B$, and $\angle BAC$ sits at $A$. This is geometry’s notation disguise, and misreading it is the most common first error. When only one angle sits at a vertex, $\angle B$ is short for it.
• Angles are measured in degrees: one full turn is $360^\circ$. An angle is acute if it is less than $90^\circ$, right if it equals $90^\circ$, obtuse if it is between $90^\circ$ and $180^\circ$, and straight if it equals $180^\circ$ (its two rays point in opposite directions and form a line). An angle more than $180^\circ$ is reflex; unless a problem says so, $\angle ABC$ means the angle that is at most $180^\circ$.
• Two angles are supplementary if they add to $180^\circ$ and complementary if they add to $90^\circ$. A ray bisects an angle if it splits it into two equal angles; the ray is the angle’s bisector.

A proof is a chain of statements in which every statement is justified by one of four things: a given (something the problem states), a definition, a postulate, or a theorem proved earlier. A postulate is a rule accepted without proof, the starting point everyone agrees on. A theorem is a statement proved from postulates and earlier theorems. Nothing else counts: not what the figure looks like, not a measurement, not “it’s obvious.”

The because test. After every line of a proof you must be able to say “because …” and finish the sentence with a given, a definition, a postulate, or a theorem by name. If you cannot, the line is not yet part of a proof.

This module starts from four postulates. You meet each one where it is first used.

We also use, without comment, the ordinary facts about position: which side of a line a point lies on, and which of two rays lies inside an angle. Euclid read these from the figure, and so do we, but only when the words of the problem guarantee them. A fuller list of postulates, David Hilbert’s from 1899, proves them too; olympiad write-ups sometimes need that care, and GEO27 returns to it.

One word from logic you will need in section 2.3. The converse of “if $P$, then $Q$” is “if $Q$, then $P$.” A statement and its converse are different claims: “if two angles are vertical angles, they are equal” is true (section 2.2), but its converse, “if two angles are equal, they are vertical angles,” is false, because two angles in different places can be equal. When both directions hold, we say “$P$ if and only if $Q$.” C04 treats converses and proof in general.

2.2. Angles at a point, and vertical angles#

Two angles are adjacent if they share a vertex and one ray and do not overlap. If the two rays they do not share point in opposite directions, the pair makes a straight angle, and P1 gives the first fact of angle chasing:

When two lines cross, they make four angles. The two angles that sit opposite each other, sharing only the vertex, are vertical angles. The figure below shows why they must be equal: each of them makes a straight angle with the same neighbour.

Proof. Number the angles $\angle1$, $\angle2$, $\angle3$, $\angle4$ in order around the crossing, so that $\angle1$ and $\angle3$ are vertical.

The same argument with $\angle3$ and $\angle4$ shows $\angle2=\angle4$.$\blacksquare$

A small example, with every step. Two lines cross. One angle measures $(2x+20)^\circ$ and the angle next to it measures $(4x-20)^\circ$. They are a linear pair, so

So the two angles are $2(30)+20=80^\circ$ and $4(30)-20=100^\circ$. By Theorem 1.1 the other two angles are also $80^\circ$ and $100^\circ$. Check: $80+100+80+100=360$, a full turn.

2.3. Parallel lines and a transversal#

Two lines in the same plane are parallel if they never meet, however far they are extended. We write $\ell\parallel m$ and mark parallel lines in figures with matching arrows. A transversal is a line that crosses two other lines at two different points. It makes four angles at each crossing, eight in all, and three kinds of pairs get names:

Here is the one assumption that makes this geometry flat.

Why assume it rather than prove it? Because it cannot be proved from P1 and the other basic facts: for two thousand years mathematicians tried, and in the 1800s they found consistent geometries in which it fails. On the surface of a sphere, for instance, any two “straight” paths (great circles) meet. Stretch S1 shows what then happens to triangles. In flat geometry, the geometry of engineering drawings, CAD, and this track, we accept it.

Everything else about parallels follows from P2.

Proof. Let $\ell$ be the upper line and $m$ the lower one, and name four angles as in the strip below: $b$ above $\ell$ and right of the transversal, $a$ below $\ell$ and left of it, and $c$ and $e$ above $m$, to the right and left of it. Then $b$ and $c$ are corresponding angles, $a$ and $c$ are alternate interior angles, and $a$ and $e$ are co-interior angles.

(a)$\Rightarrow$(b).

If $\ell\parallel m$, then $b=c$ (P2). Also $a=b$, because they are vertical angles (Theorem 1.1). So $a=c$.

(b)$\Rightarrow$(a).

If $a=c$, then $b=a=c$ (vertical angles again), so the corresponding angles $b$ and $c$ are equal, and P2 says $\ell\parallel m$.

(b)$\Leftrightarrow$(c).

The angles $c$ and $e$ form a linear pair, so $c+e=180^\circ$ (P1). Hence $a=c$ exactly when $a+e=180^\circ$.

The other alternate pair and the other co-interior pair work the same way on the other side of the transversal.$\blacksquare$

One more consequence, which worked example 1 needs:

Proof. Suppose $k$ and $m$ were different lines that met at a point $X$. Then $X$ would not lie on $\ell$ (both $k$ and $m$ miss $\ell$), and $k$ and $m$ would be two different lines through $X$ parallel to $\ell$. That contradicts the “exactly one” in P2. So $k$ and $m$ never meet: they are parallel.$\blacksquare$

This is a proof by contradiction: assume the opposite of what you want, and show that it breaks something already accepted.

Theorem 1.2 works in both directions, and you will use both. Forwards: the lines are given parallel, so angles are equal. Backwards: you have shown two angles are equal, so the lines are parallel. The second direction is the only way to prove lines parallel in this module.

2.4. The triangle angle sum#

A triangle $ABC$ is three points not on one line, its vertices, joined by three segments, its sides. Its interior angles are $\angle BAC$, $\angle ABC$, and $\angle BCA$, written $\angle A$, $\angle B$, and $\angle C$ for short. The tearing picture in section 1 suggests they always add to $180^\circ$. Here is why, as a proof.

Proof. Write $\alpha=\angle A$, $\beta=\angle B$, $\gamma=\angle C$.

$\blacksquare$

Three consequences follow at once, and you will use them without comment from now on:

• A triangle has at most one right or obtuse angle (two of them would already add to $180^\circ$ or more).
• In a right triangle, the two other angles add to $90^\circ$: they are complementary.
• If two angles of one triangle equal two angles of another, the third angles are equal too.

A small example, with every step. A triangle’s angles are $2x$, $3x$, and $4x$. By Theorem 1.3, $2x+3x+4x=180$, so $9x=180$ and $x=20$. The angles are $40^\circ$, $60^\circ$, and $80^\circ$. Check: $40+60+80=180$.

2.5. The exterior angle theorem#

Extend side $BC$ of triangle $ABC$ past $C$ to a point $D$. The angle $\angle ACD$, between side $CA$ and the extension, is an exterior angle of the triangle at $C$. It forms a linear pair with the interior angle $\angle ACB$. The other two interior angles, $\angle A$ and $\angle B$, are its remote interior angles: the ones not touching it.

Proof.

$\blacksquare$

The exterior angle theorem is the angle chaser’s shortcut: it gives an angle in one step that would otherwise take a triangle sum and a linear pair. One consequence: an exterior angle is larger than either remote interior angle.

2.6. Isosceles triangles#

A triangle is isosceles if at least two of its sides are equal, equilateral if all three are, and scalene if no two are. In an isosceles triangle the two equal sides are called its legs, the third side its base, the angle between the legs its apex angle, and the two angles at the ends of the base its base angles. (Careful: “leg” has a second meaning. In a right triangle, the legs are the two sides that form the right angle, and the side opposite the right angle is the hypotenuse. The context tells you which is meant.)

To prove anything about equal sides, we need a rule that connects sides to angles. Two triangles are congruent if one can be moved onto the other exactly, by sliding, turning, or flipping it over: then every side and every angle of one equals the matching side and angle of the other. We write $\triangle ABC\cong\triangle DEF$, listing matching vertices in the same order.

GEO02 takes SAS as its starting postulate and proves ASA from it without using anything in this section, so borrowing both here is not circular. (GEO02 will use Theorem 1.5 to prove a third rule, SSS; that direction of dependence is fine, because Theorem 1.5 rests on SAS alone.)

Proof (an ancient argument usually credited to Pappus of Alexandria). Compare triangle $ABC$ with the same triangle read in a different order, $ACB$, as if it had been flipped over.

For the converse, compare $ABC$ with $ACB$ again: $\angle B=\angle C$ (given), $BC=CB$, and $\angle C=\angle B$, so $\triangle ABC\cong\triangle ACB$ by ASA (P4), and the matching sides give $AB=AC$.$\blacksquare$

Two more facts follow. An equilateral triangle has all three angles equal (apply Theorem 1.5 twice), and so each is $180^\circ/3=60^\circ$. And in an isosceles triangle with apex angle $a$, each base angle is $\tfrac{180^\circ-a}{2}=90^\circ-\tfrac a2$.

2.7. Polygons#

A polygon is a closed chain of segments, its sides, joined end to end at its vertices, with no two sides crossing. A polygon with $n$ sides is an $n$-gon: a triangle has 3, a quadrilateral 4, a pentagon 5, a hexagon 6, a heptagon 7, an octagon 8. A diagonal joins two vertices that are not neighbours. A polygon is convex if each interior angle is less than $180^\circ$; then every diagonal lies inside it. A polygon is regular if all its sides are equal and all its angles are equal. At each vertex, extending one side makes an exterior angle, which forms a linear pair with the interior angle there.

Proof. (a) Pick one vertex and draw the diagonals from it to every vertex except itself and its two neighbours: $n-3$ diagonals. They cut the polygon into $n-2$ triangles. Because the polygon is convex, the diagonals lie inside it and split the angles at the chosen vertex (P1), so the triangles’ angles together make up exactly the polygon’s angles. Each triangle contributes $180^\circ$ (Theorem 1.3), so the total is $(n-2)\cdot180^\circ$. (b) At each of the $n$ vertices, interior plus exterior is $180^\circ$ (linear pair), so all $n$ pairs total $n\cdot180^\circ$. Subtract the interior total from (a): $n\cdot180^\circ-(n-2)\cdot180^\circ=360^\circ$.$\blacksquare$

Part (b) has a picture you can act out: walk around the polygon’s edge. At each corner you turn through the exterior angle, and after one lap you face the way you started, having turned exactly once around. A car driving once around a convex block turns through $360^\circ$ in total, however the block is shaped, so a robot that adds up its heading changes around a closed loop has a built-in check: the total must come to one full turn. (For a non-convex polygon, part (a) still holds with a different way of cutting it into triangles, and part (b) holds if a turn the other way counts as negative. Stretch S2 explores both.)

A small example, with every step. A regular pentagon has interior angle sum $(5-2)\cdot180^\circ=540^\circ$, so each of its five equal angles is $540^\circ/5=108^\circ$. Each exterior angle is $360^\circ/5=72^\circ$. Check: $108+72=180$, a linear pair.

2.8. Angle chasing: the method#

Angle chasing is how you find an unknown angle, or prove two angles equal, by passing known angles through a figure one fact at a time. It is bookkeeping, and it rewards the same discipline as the Modular Workbench in calculus: set out the parts on separate lines, check each one, then assemble.

1. Copy the figure large, by hand, and mark every given: tick marks, arcs, parallel arrows, right-angle squares, and known measures.
2. Guess. Estimate the target from the figure and write the estimate down. In this track the figures are drawn exactly unless they say “not to scale,” so a careful estimate should land within a few degrees. The guess is not evidence; it is a check on your proof and a way to train your eye.
3. Name one unknown. Call the target $x$, or call $x$ an angle that many others depend on.
4. Push facts outward, one per line. For each new angle, write its value and its reason: “$\angle ADB=2x$ (exterior angle of triangle $ACD$).” This is the Workbench: one angle, one reason, one line.
5. Close the loop. When some angle can be found two ways, or a triangle’s three angles are all known in terms of $x$, write the equation and solve it.
6. Check. Put the value back: does every triangle add to $180^\circ$, every straight line to $180^\circ$? How far was your guess?
7. Write the proof (section 2.9).

There are two styles of chase, and good solvers use both. A synthetic chase uses the theorems as moves and names each one. An algebraic chase gives letters to several angles and writes one equation per fact (each triangle, each straight line, each isosceles triangle, each parallel pair), then solves the system. The algebraic chase always works when the figure’s angles are determined by the givens; it is how a computer would do it (stretch S3). The synthetic chase is shorter when you see the right move, and it explains why. Worked examples 1 and 2 solve the same problem both ways.

2.9. Writing it up#

A proof is written for a reader who was not in the room: a grader, an interviewer, or you in three weeks. Every proof in this track has the same shape.

Proofs come in two layouts. A two-column proof, like the ones above, puts statements on the left and reasons on the right; it is the best way to learn, because a missing reason is a visible hole. A paragraph proof says the same thing in sentences (“Since $\ell\parallel m$, the alternate angles $\angle1$ and $\angle2$ are equal by Theorem 1.2”); it is how proofs are written in olympiads and papers. Here is Theorem 1.4 as a paragraph: Extend $BC$ past $C$ to $D$. The angles $\angle ACD$ and $\angle ACB$ form a linear pair, so they add to $180^\circ$; by the triangle angle sum, $\angle A+\angle B$ adds with $\angle ACB$ to $180^\circ$ too. Hence $\angle ACD=\angle A+\angle B$.$\;\blacksquare$

Four rules for every write-up:

1. Every step has a reason, and a theorem is cited by its name or number.
2. Never write “by the figure.” Positions (which side, which order) may be read from a figure the words guarantee; measures, equalities, midpoints, and parallels may not.
3. Say which triangle you apply a theorem to: “angle sum in triangle $ABD$.”
4. Introduce every auxiliary line with a sentence that says why it exists: “Draw the line through $C$ parallel to $AB$ (P2).”

Circular reasoning is using the thing you are proving, directly or through a chain. Theorems in this module depend on each other in a fixed order:

What proves what in GEO01. A proof of any theorem may use only what sits below it.

A proof may use only what sits below its theorem in that map. Proving the angle sum from the exterior angle theorem is circular here, because the exterior angle theorem was proved from the angle sum. Problem GEO01.20 asks you to find exactly this flaw.

A proof by example is not a proof. Measuring one figure, or checking three triangles, shows that a claim is plausible. A proof must work for every figure the words allow.

2.10. You cannot measure a diagram#

Three figures to train your suspicion. The first two are drawn exactly and still mislead the eye; the third is deliberately not to scale, as contest figures sometimes are.

In this track every figure is exact unless it says “not to scale,” so a guess from the figure is a fair estimate. That is what makes the guess-then-prove habit useful: a guess far from your answer is a signal to recheck the work. But a guess is never a reason in a proof.

2.11. Beginner mistakes: a field guide#

Each mistake below has a name, so that whoever marks your work, you included, can say which one it was, and so that your error log can count them.

3. Worked examples#

Each example starts with a guess. Cover the solution, write your estimate, and try the problem for five minutes before reading on.

3.1. Worked example 1: the bent arrow#

Problem. Lines $\ell$ and $m$ are parallel. Point $A$ is on $\ell$, point $B$ is on $m$, and point $P$ lies between the lines, to the right of both. The angle between $AP$ and the part of $\ell$ to the right of $A$ is $38^\circ$, and the angle between $BP$ and the part of $m$ to the right of $B$ is $47^\circ$. Find $\angle APB$.

Approach A: a parallel through the bend. The angles we know sit at $\ell$ and at $m$, far from $P$. A parallel through $P$ can carry both of them to $P$.

Approach B: make a triangle. Instead of a new parallel, extend a line we already have until it closes a triangle, then use the exterior angle theorem.

Answer: $\angle APB=\boxed{85^\circ}$. How close was your guess?

Comparing the approaches. Approach B is one step shorter because the exterior angle theorem packs a triangle sum and a linear pair into one move. Approach A generalizes: a path that bends many times between two parallels can be split by a parallel through every bend, which gives the zigzag rule of GEO01.13. Both show the same pattern: the angle at the bend is the sum of the two outer angles. Approach B also shows a habit worth keeping: when an angle has no triangle around it, extend a line and make one.

3.2. Worked example 2: an isosceles chain#

Problem. In triangle $ABC$, point $D$ lies on side $BC$ with $AD=BD$ and $CA=CD$. If $\angle B=25^\circ$, find $\angle BAC$.

Approach A: chase forward from the given angle. Each isosceles triangle turns one known angle into another.

Approach B: let the given angle be a letter. Write $\angle B=x$ instead of $25^\circ$ and run the same chase: $\angle BAD=x$, $\angle ADC=2x$, $\angle CAD=\angle CDA=2x$, so

Check with the angle sum of triangle $ABC$: $x+3x+(180^\circ-4x)=180^\circ$ for every $x$, as it must. With $x=25^\circ$: $\angle BAC=\boxed{75^\circ}$ and $\angle C=80^\circ$.

Comparing the approaches. Approach A is the quickest route to the number. Approach B costs one extra line and pays three times over: it shows that $\angle BAC$ is always three times $\angle B$ in this figure; it shows the figure exists only when $180^\circ-4x>0$, that is, when $\angle B<45^\circ$; and the angle sum of triangle $ABC$ becomes a check rather than a step. When a problem gives you one number, try replacing it with a letter: the general answer often explains the specific one.

3.3. Worked example 3: a proof, and a non-proof#

Problem. Lines $\ell$ and $m$ are parallel, and a transversal meets $\ell$ at $A$ and $m$ at $B$. The two co-interior angles on the right of $AB$ are bisected, and the two bisectors meet at $P$. Prove that $\angle APB=90^\circ$.

Given: $\ell\parallel m$; $A$ on $\ell$ and $B$ on $m$; ray $AP$ bisects the co-interior angle at $A$ on the right of $AB$, and ray $BP$ bisects the co-interior angle at $B$ on the right of $AB$. To prove: $\angle APB=90^\circ$. Plan: the two whole angles add to $180^\circ$, so their halves add to $90^\circ$, and triangle $APB$ gets the rest.

Proof. Let the co-interior angle at $A$ be $2x$ and the one at $B$ be $2y$.

$\blacksquare$

A non-proof, and what is wrong with it. A student writes: “I drew the figure carefully, measured $2x=64^\circ$ and $2y=116^\circ$, and found $\angle APB=90^\circ$. So the angle is always $90^\circ$.” Every number is correct, and the conclusion happens to be true, but this is not a proof. It checks one figure; the claim is about every figure the words allow (mistake M9). The measurement also brings in protractor error (mistake M1). The student’s work is good evidence, which is exactly what a guess is for, but only the five steps above make it certain.

The paragraph version. Since $\ell\parallel m$, the co-interior angles at $A$ and $B$ add to $180^\circ$, so their halves $\angle PAB$ and $\angle PBA$ add to $90^\circ$. By the angle sum in triangle $APB$, $\angle APB=180^\circ-90^\circ=90^\circ$.$\;\blacksquare$

3.4. Worked example 4: the five points of a star#

Problem. Five points $A$, $B$, $C$, $D$, $E$ are the corners, in order, of a convex pentagon of any shape. Draw the five-pointed star $A\to C\to E\to B\to D\to A$. Prove that the five angles at its tips add to $180^\circ$.

Approach A: two exterior angles and one triangle. Name the crossings on the line $EB$: let it meet $AD$ at $G$ and $AC$ at $F$, as in the figure. Tip $A$ is an angle of the small triangle $AGF$, and each of the other two angles of that triangle is an exterior angle of a triangle whose remote angles are two of the other tips.

$\blacksquare$

Approach B: turn a pencil. Lay a pencil along side $AC$ at tip $A$, pointing toward $C$. Slide it to $C$ and turn it through tip $C$’s angle so that it lies along $CE$; slide it to $E$ and turn it through tip $E$’s angle onto $EB$; and so on around the star, back to $A$. Every turn is in the same direction, and when you arrive back along $AC$ the pencil points the opposite way: it has turned through half a revolution, $180^\circ$, and the turns were exactly the five tip angles. This is a convincing picture rather than a complete proof as written (one must check that the total is $180^\circ$ and not $180^\circ$ plus a full turn), but it explains the number, and challenge GEO01.22 turns the same idea into a proof for bigger stars.

Comparing the approaches. Approach A is a complete proof built from this module’s theorems, and it is the one to write up. Approach B is the mechanism: the star’s tips are the turns of a path that ends up reversed. Engineers meet the same bookkeeping when they add up a vehicle’s heading changes around a route.

3.5. Worked example 5: which regular polygons tile a floor?#

Problem. Copies of one regular polygon are to cover a floor with no gaps and no overlaps, meeting corner to corner. Which regular polygons can do it?

Solution. Where corners meet, the angles around the point must add to exactly $360^\circ$, so the interior angle of the polygon must divide $360^\circ$ exactly. A regular $n$-gon’s interior angle is $\dfrac{(n-2)\cdot180^\circ}{n}$ (Theorem 1.6).

• $n=3$: $60^\circ$, and $360/60=6$ triangles meet at each point.
• $n=4$: $90^\circ$, and 4 squares meet.
• $n=5$: $108^\circ$, and $360/108=3\tfrac13$: three pentagons leave $360-324=36^\circ$, four overlap.
• $n=6$: $120^\circ$, and 3 hexagons meet.
• $n\ge7$: the interior angle is more than $120^\circ$ (it grows with $n$) and less than $180^\circ$, so $360$ divided by it lies strictly between 2 and 3, never a whole number.

Answer: only equilateral triangles, squares, and regular hexagons. That is why tile-based game maps use squares, hexagons, or triangles, and never regular pentagons. Challenge GEO01.24 asks what happens when three different regular polygons share a corner.

4. Practice#

Three tiers. Drill practises one move until it is automatic. Problems need several moves or a written proof. Challenges need an idea you have to find. Every problem has a three-rung hint ladder after the practice set (the [hints] link jumps there): rung 1 is a question, rung 2 points to the idea, rung 3 is a skeleton with blanks. Climb one rung at a time and try again after each. The full solutions, with figures, are in a separate section at the end of the chapter; open one only after you have written your own attempt.

Before every problem with a figure: guess. Write your estimate beside the problem number, then solve, then write the answer and the difference. Keep the column: your guesses should get sharper over the module, and a big gap is a signal to recheck your work.

4.1. Drill#

4.2. Problems#

4.3. Challenges#

5. Toolbox cards#

One card per theorem, one page each, to review like flashcards: cover everything except the title, say the statement aloud, sketch the diagram, then uncover and check. Print this section, or flip through it on screen.

Toolbox card 1.1Straight angles and vertical angles

Statement

Angles on a straight line add to $180^\circ$ (a linear pair); angles all the way around a point add to $360^\circ$. When two lines cross, vertical angles are equal (Theorem 1.1).

Reach for it when

Two lines cross anywhere in the figure; an angle sits next to a straight line; you need a second angle at a crossing where you know one.

Watch out

The two angles of a linear pair must share a ray, and their outer rays must form one straight line. Two angles that merely touch do not add to $180^\circ$.

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

Toolbox card 1.2Parallel lines and a transversal

Statement

If a transversal crosses two parallel lines, corresponding angles are equal (F), alternate interior angles are equal (Z), and co-interior angles add to $180^\circ$ (C). Conversely, any one of those three facts proves the lines parallel (P2 and Theorem 1.2). Two lines parallel to the same line are parallel to each other (Corollary 1.2a).

Reach for it when

The words or arrows say two lines are parallel (use it forwards); or you must prove two lines parallel (use it backwards).

Watch out

Never use it forwards unless the parallel is given or already proved (M3). F and Z pairs are equal; the C pair is not equal, it adds to $180^\circ$ (M5).

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

Toolbox card 1.3The triangle angle sum

Statement

In every triangle, $\angle A+\angle B+\angle C=180^\circ$ (Theorem 1.3). Proof: the parallel to $AB$ through $C$ carries $\angle A$ and $\angle B$ to $C$, beside $\angle C$ on a straight line. Consequences: at most one right or obtuse angle; the acute angles of a right triangle add to $90^\circ$; two equal pairs of angles force the third pair equal.

Reach for it when

You know two angles of a triangle, or all three in terms of one unknown; you need to close the loop of an angle chase.

Watch out

It holds in the flat plane only; it depends on the parallel postulate. Name the triangle you apply it to.

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

Toolbox card 1.4The exterior angle theorem

Statement

An exterior angle of a triangle equals the sum of the two remote interior angles: $\angle ACD=\angle A+\angle B$ (Theorem 1.4). It is bigger than either of them.

Reach for it when

An angle sits outside a triangle along an extended side; you want one step instead of an angle sum plus a linear pair; an angle has no triangle around it (extend a line and make one).

Watch out

Remote means not touching: the interior angle at the same vertex is not part of the sum (M6).

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

Toolbox card 1.5The isosceles triangle

Statement

If $AB=AC$, then $\angle B=\angle C$; conversely, if $\angle B=\angle C$, then $AB=AC$ (Theorem 1.5, from SAS and ASA). Each base angle is $90^\circ-\tfrac12(\text{apex angle})$. An equilateral triangle has three $60^\circ$ angles.

Reach for it when

Two equal sides share a vertex (tick marks, or two radii of one circle in Phase 2); you need to turn an angle fact into a side fact or back.

Watch out

The equal angles sit opposite the equal sides, at the ends of the base, not at the apex. Equal only if given or proved, never because they look equal (M2).

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

Toolbox card 1.6Polygon angle sums

Statement

A convex $n$-gon’s interior angles add to $(n-2)\cdot180^\circ$, and its exterior angles, one per vertex, add to $360^\circ$ (Theorem 1.6). A regular $n$-gon has interior angle $\tfrac{(n-2)\cdot180^\circ}{n}$ and exterior angle $\tfrac{360^\circ}{n}$.

Reach for it when

Any polygon with more than three sides; regular polygons; tilings; a path that turns at corners and comes back to its start.

Watch out

The exterior sum is $360^\circ$ for every $n$, and the interior sum grows with $n$. For non-convex polygons, count turns the other way as negative.

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

Toolbox card 1.7The angle chase and the write-up

Statement

Copy the figure large and mark the givens. Guess. Name one unknown $x$. Push facts outward one per line, each with its reason (the Workbench). Close the loop with an equation. Check every triangle and line sums correctly, and compare with the guess. Then write: Given, To prove, Plan, numbered steps with reasons, $\blacksquare$.

Reach for it when

Every angle problem. When stuck: extend a line to make a triangle, draw a parallel through a bend, or replace a given number with a letter.

Watch out

The because test on every line. No step from the figure alone (M1, M2, M3), no circular steps (M4), and no proof by example (M9).

Review this card at 2, 7, and 21 days: cover everything but the title, say the statement and draw the diagram, then check.

6. The mastery gate#