# Why the angles of a triangle add to 180°

> Tear the corners off any triangle and they line up along a straight edge. A single parallel line explains why, and a sphere shows the line is needed.

Series: Geometry (https://www.monocurl.com/monocurl-essays/geometry/)
Canonical: https://www.monocurl.com/monocurl-essays/geometry/triangle-angle-sum/
Keywords: triangle angle sum, angles of a triangle add to 180, triangle angle sum proof, exterior angles of a polygon, interior angles of a polygon, alternate angles

[Figure 1, animation: The three corners of a paper triangle are torn off and pushed together. Their outer edges form one straight line.]

Cut any triangle out of paper, tear off its three corners and push the three points together, edge against edge. The outer edges of the pieces form a straight line, as in the animation above, whatever triangle you started with.

Nobody chose those corners to fit together, and changing the shape of the triangle changes all three. So why should they add up to a straight line? And is the line exactly straight, or only so close that torn paper can't show the difference?

## What the experiment shows

Angles are measured in **degrees**, written °, with a full turn divided into 360 of them. Half a turn is then 180°, the angle along one side of a straight line, and a quarter turn is 90°, a **right angle**. So the paper experiment says that the three angles of the triangle add to 180°.

The triangle in the animation has angles of 50°, 60° and 70°, and indeed $50 + 60 + 70 = 180$. One triangle proves nothing, so the next figure tries more. The base stays where it is, the top corner moves, and each time the three angles are measured from the drawing and copies of them are laid side by side.

[Figure 2, steps: Four different triangles. In each one the three angles are measured, copied, and placed side by side. Every time they fill exactly half a turn.]

  - step: Angles of 50°, 70° and 60° fill half a turn.
  - step: A right triangle: 90°, 55° and 35°.
  - step: A triangle with an obtuse angle: 30°, 110° and 40°.
  - step: A long flat triangle: 25°, 135° and 20°.
All four sums are 180°. Watch what happens between the steps: when the top corner is pulled down toward the base, the two base angles shrink and the top angle opens by the same amount, so the total stays put.

## Why measuring cannot settle the question

Measurement has two weaknesses that no amount of care removes. A protractor reads to about half a degree, and torn paper is worse, so if the true sum were 179.9°, or 180.3° for long thin triangles, our experiments would not notice. And there are infinitely many triangles, of which we have tried four. We want a reason that covers every triangle at once and leaves no room for an error of a tenth of a degree.

## Bringing the corners to one place

In the opening animation the corners were carried off to a spare line at the side, a place that has nothing to do with the triangle and so is hard to reason about. It is tidier to leave one corner where it is and bring the other two to it. Name the corners $A$, $B$ and $C$, with $C$ at the top, and keep $C$ fixed. We want a copy of the angle at $A$ to sit at $C$, right beside the angle that is already there, sharing the side $CA$ with it.

Sliding the copy up along the side won't do, because a slid copy still opens toward the right, the way the original does at $A$, and at $C$ it would sit above the triangle. The copy has to be turned around. The motion that does this is a **half turn**: a rotation through 180° about some center point. A half turn about a center $M$ sends each point to the point directly opposite it through $M$, at the same distance. If we want $A$ to land on $C$, the center must be the point halfway between them, the midpoint of the side $AC$.

**Pause and try.** The angle at $A$ has two edges. One runs along the side $AC$, and the other runs along the base, pointing to the right. Give the whole corner a half turn about the midpoint of $AC$. Where does each edge end up, and which way does it point?

**Answer.** The half turn swaps $A$ and $C$, so the edge that ran from $A$ toward $C$ now runs from $C$ toward $A$ and still lies along the side of the triangle. The other edge started at $A$ pointing right, along the base. A half turn reverses every direction, so this edge now starts at $C$ and points left, level with the base but up at the top corner.

Two lines in a plane that never meet are called **parallel**. A line and its half-turned copy never meet (a property of half turns listed below), so that second edge lies along a line through $C$ parallel to the base. The same thing happens on the other side: a half turn about the midpoint of $BC$ carries the corner at $B$ up to $C$, and its base edge ends up starting at $C$ and pointing right, again along a line through $C$ parallel to the base.

After both half turns three angles sit at $C$, and whether they make a straight line comes down to one question. The edge pointing left and the edge pointing right each lie on a line through $C$ parallel to the base, and we need those to be the *same* line. If only one line through $C$ is parallel to the base, the two edges are its two halves, and the three angles fill one side of a straight line exactly.

### The fact about parallel lines that we need

The step that carried the angle from $A$ to $C$ has nothing to do with triangles, so let us isolate it. Take two parallel lines $\ell$ and $m$, and a third line $t$ that crosses both. The figure below gives the picture a half turn about the point $M$ halfway between the two crossings.

[Figure 3, animation: Two parallel lines \ell and m are crossed by a line t at P and Q. A half turn about the midpoint M of PQ carries \ell onto m and carries the blue angle at P onto the angle at Q on the other side of t.]

The two angles that end up matched are called **alternate angles**: they lie between the two parallel lines, at the two crossings, on opposite sides of the crossing line. The proof that they are equal uses three properties of a half turn, which I'll list first so that nothing is hidden.

**Note.** **What we assume about a half turn.** Let $H$ be the half turn about a point $M$.

- $H$ is a rigid motion: it sends lines to lines and keeps every length and every angle the same.
- For every point $X$, the point $M$ is the midpoint of $X$ and $H(X)$. So $H$ swaps the two ends of any segment whose midpoint is $M$, and it sends every line through $M$ to itself.
- $H$ sends a line that does not pass through $M$ to a line parallel to it.

All three can be proved from Euclid's axioms about congruent triangles without the parallel axiom below. (Property 3 is Proposition 27 of Book I of the *Elements*, applied to a line and its half-turned copy.) So a half turn shows that a parallel through a given point *exists*. Whether it is the only one is a separate matter.

**The parallel axiom.** Through a point not on a line, there is exactly one line parallel to the given line.

**Theorem.** **(Alternate angles.)** Let $\ell$ and $m$ be parallel lines in the plane, and let a line $t$ cross $\ell$ at $P$ and $m$ at $Q$. Let $X$ be a point of $\ell$ other than $P$, and let $Y$ be a point of $m$ other than $Q$, with $X$ and $Y$ on opposite sides of $t$. Then $\angle XPQ = \angle YQP$.

**Proof.** Let $M$ be the midpoint of the segment $PQ$ and let $H$ be the half turn about $M$. By property 2, $H$ swaps $P$ and $Q$ and sends the line $t$ to itself.

The line $\ell$ misses $M$, since otherwise it would contain $P$ and $M$, would be the line $t$, and would meet $m$ at $Q$. So by property 3, $H(\ell)$ is a line parallel to $\ell$, and it passes through $H(P) = Q$. The line $m$ is also parallel to $\ell$ and passes through $Q$. By the parallel axiom there is only one such line, so $H(\ell) = m$.

The point $X$ is not on $t$, and by property 2 the segment from $X$ to $H(X)$ crosses $t$ at $M$, so $X$ and $H(X)$ lie on opposite sides of $t$. Thus $H(X)$ is a point of $m$ on the side of $t$ that contains $Y$, which means that $H(X)$ lies on the ray from $Q$ through $Y$.

Therefore $H$ carries the ray $PX$ onto the ray $QY$ and the ray $PQ$ onto the ray $QP$, so it carries the angle $\angle XPQ$ onto the angle $\angle YQP$. By property 1 the two angles are equal.

The parallel axiom entered the proof at one point, the sentence "there is only one such line". Without it, $H(\ell)$ and $m$ could be two different parallels through $Q$, and the angles would have no reason to match.

## The proof for a triangle

The steps below carry out the plan on the 50°, 60°, 70° triangle. The new line is the one line through $C$ parallel to the base, and each half-turned copy lands against it for the reason just proved.

[Figure 4, steps: The proof. A line through the top corner parallel to the base receives half-turned copies of the two base angles, and the three angles then fill one side of that line.]

  - step: A triangle with angles of 50°, 60° and 70°.
  - step: Draw the line through C that is parallel to the base AB.
  - step: Give a copy of the blue corner a half turn about M, the midpoint of AC. It lands at C with one edge along the new line.
  - step: Give a copy of the orange corner a half turn about N, the midpoint of BC. It lands on the other side of C.
  - step: At the top, the three angles fill one side of a straight line. They add to 180°.
Here is the same argument written out for an arbitrary triangle.

**Theorem.** **(Angle sum of a triangle.)** Let $ABC$ be a triangle in the plane. Then

$$
\angle CAB + \angle ABC + \angle BCA = 180°.
$$

**Proof.** Let $\ell$ be the line through $C$ parallel to the line $AB$. It exists and there is only one, by the parallel axiom. Choose a point $D$ on $\ell$ on the opposite side of the line $AC$ from $B$, and a point $E$ on $\ell$ on the opposite side of the line $BC$ from $A$. Then $D$ and $E$ lie on opposite sides of $C$ along $\ell$.

The line $AC$ crosses the parallel lines $AB$ and $\ell$ at $A$ and $C$, and the points $B$ and $D$ are on opposite sides of it. By the alternate angles theorem, $\angle DCA = \angle CAB$.

In the same way, the line $BC$ crosses the parallel lines $AB$ and $\ell$ at $B$ and $C$, with $A$ and $E$ on opposite sides of it, so $\angle ECB = \angle ABC$.

Now look at the point $C$. The line $AB$ never meets $\ell$, so $A$ and $B$ lie on the same side of $\ell$, and the rays $CA$ and $CB$ both point into that side. Going around $C$ from the ray $CD$ to the ray $CE$ on that side, we pass the ray $CA$ and then the ray $CB$. So the three angles $\angle DCA$, $\angle ACB$ and $\angle BCE$ sit side by side without overlapping and together make up the straight angle $\angle DCE$:

$$
\angle DCA + \angle ACB + \angle BCE = 180°.
$$

Replacing the first and last angles by their equals gives $\angle CAB + \angle BCA + \angle ABC = 180°$.

One kind of step was taken from the picture: the statements about which side of a line a point lies on, and the order of the four rays around $C$. A fully formal treatment derives them from axioms about order, which I omit.

This answers both opening questions. The corners fit because two of them are alternate angles to the pieces of a straight line through the third. And the fit is exact: the proof contains no measurement, so there is no tenth of a degree for an error to hide in.

**Key idea.** The three angles of a triangle add to 180° because a line through one corner, parallel to the opposite side, is split at that corner into copies of all three angles.

Besides those facts about order, two things were assumed: the three properties of a half turn, which can be proved without the parallel axiom, and the parallel axiom itself, which cannot be proved from Euclid's other axioms. The last section shows what happens without it.

### Using the theorem

**Worked example.** **Angles given by expressions.** The angles of a triangle are $x + 10$, $2x$ and $3x - 40$ degrees. Find all three.

We don't know any angle yet, but we know their total, and that gives one equation in the one unknown:

$$
(x + 10) + 2x + (3x - 40) = 180.
$$

The left side collects to $6x - 30$, so $6x = 210$ and $x = 35$. The angles are $35 + 10 = 45°$, $2 \cdot 35 = 70°$ and $3 \cdot 35 - 40 = 65°$.

Check: $45 + 70 + 65 = 180$, and all three are positive, as the angles of a triangle must be. The second check matters, because an equation like this can have a solution that makes one "angle" zero or negative, and then no such triangle exists.

### The angle outside a corner

Extend one side of a triangle past a corner. The angle between the extension and the other side at that corner is called an **exterior angle** of the triangle. The angles inside are **interior angles** when we need to tell them apart.

[Figure 5, image: The base of the 50°, 60°, 70° triangle is extended past B to D. The exterior angle at B is 120°, and the line through B parallel to AC splits it into copies of the angles at A and C.]

In the figure the exterior angle at $B$ is 120°, and the two interior angles away from $B$ are 50° and 70°, which add to 120°. The reason takes one line. The exterior angle and the interior angle at $B$ lie along a straight line, so the exterior angle is $180° - \angle B$. By the theorem, $180° - \angle B$ is also $\angle A + \angle C$.

**Theorem.** **(Exterior angle.)** An exterior angle of a triangle equals the sum of the two interior angles at the other two corners.

## A second view: walking around the triangle

There is another way to see the number 180, and it explains what happens for shapes with more sides. Imagine walking along the edge of the triangle, counterclockwise, starting in the middle of the base. At each corner you have to turn left to follow the next side. The amount you turn is exactly the exterior angle at that corner: before turning you face along the extension of the side you were on, and after turning you face along the next side.

[Figure 6, animation: A walker goes once around the triangle. At each corner the turn is the green angle outside the triangle, and the dial adds up the turning to a full 360°.]

On the 50°, 60°, 70° triangle the walker turns through $180° - 60° = 120°$ at the first corner, then 110°, then 130°, and the dial fills up completely: $120 + 110 + 130 = 360$. The walker ends the trip facing the way it started, having turned left all the way and never right, so the total turning is one full turn.

This gives the angle sum again. If the interior angles are $\alpha$, $\beta$ and $\gamma$, the three turns are $180° - \alpha$, $180° - \beta$ and $180° - \gamma$, and they total 360°:

$$
(180° - \alpha) + (180° - \beta) + (180° - \gamma) = 360°.
$$

The left side is $540° - (\alpha + \beta + \gamma)$, so $\alpha + \beta + \gamma = 180°$.

This argument is a sketch. It rests on the claim that a walker who goes once around and returns to the same heading has turned through exactly 360°, and a careful proof of that needs to compare directions at different points of the plane, which is what parallel lines are for. Below, the 360° total is proved from the triangle theorem. Meanwhile the walker gives us a prediction, because nothing in the story cared that the path had three corners.

## From triangles to polygons

A **polygon** is a closed shape whose boundary is made of straight sides: 3 for a triangle, 4 for a quadrilateral, 5 for a pentagon. I'll consider **convex** polygons, the ones with no dents. Precisely, a polygon is convex when the segment joining any two of its corners stays inside it.

**Pause and try.** A walker goes once around a convex pentagon, turning left at each of its 5 corners, and turns 360° in total. Each turn is 180° minus the interior angle at that corner. Use this to predict the sum of the five interior angles.

**Answer.** The five turns add to $5 \cdot 180°$ minus the sum of the interior angles, and they total 360°. So the interior angles add to $5 \cdot 180° - 360° = 900° - 360° = 540°$.

To prove such a formula we can lean on the triangle theorem. Pick one corner of the polygon and draw a segment from it to every corner it is not already joined to. These segments are called **diagonals**, and they cut the polygon into triangles.

[Figure 7, image: A quadrilateral, a pentagon and a hexagon, each cut into triangles by lines from one corner. A shape with n sides always gives n - 2 triangles.]

The count of triangles goes 2, 3, 4 for 4, 5, 6 sides. In general there are $n - 2$ of them. Each triangle uses one side of the polygon as the side opposite the chosen corner, and the two sides of the polygon that touch the chosen corner are the only ones not used this way.

**Theorem.** **(Angle sum of a convex polygon.)** The interior angles of a convex polygon with $n$ sides, $n \ge 3$, add to $(n - 2) \cdot 180°$.

**Proof.** Name the corners $V_1, V_2, \dots, V_n$ in order around the polygon. Because the polygon is convex, each diagonal from $V_1$ lies inside it, and the diagonals $V_1V_3, V_1V_4, \dots, V_1V_{n-1}$ divide it into the $n - 2$ triangles $V_1V_2V_3$, $V_1V_3V_4$, $\dots$, $V_1V_{n-1}V_n$, which do not overlap.

Every angle of every triangle is a piece of an interior angle of the polygon. At $V_1$ the triangles' angles sit side by side and fill the polygon's angle there. At each of $V_3, \dots, V_{n-1}$ two triangles meet and their angles together fill the polygon's angle. At $V_2$ and $V_n$ a single triangle's angle is the polygon's angle. So the sum of the polygon's interior angles equals the sum of all the angles of all the triangles, which is $(n - 2) \cdot 180°$ by the triangle theorem.

For $n = 5$ this gives $3 \cdot 180° = 540°$, the walker's prediction. The formula is also true for polygons with dents, but this proof does not cover them, because a diagonal from a badly chosen corner can pass outside the shape.

We can now settle the walker's claim for every convex polygon. At each corner the exterior angle is 180° minus the interior angle, so the $n$ exterior angles, one per corner, add to

$$
n \cdot 180° - (n - 2) \cdot 180° = 2 \cdot 180° = 360°.
$$

**Theorem.** **(Exterior angles.)** The exterior angles of a convex polygon, one at each corner, add to 360°, whatever the number of sides.

**Worked example.** **Which regular polygon has 150° corners?** A polygon is **regular** when all its sides are equal and all its angles are equal. Find the number of sides of a regular polygon whose interior angles are each 150°.

The quick route is through the exterior angles, because their total does not depend on $n$. Each exterior angle is $180° - 150° = 30°$, there are $n$ of them, and they add to 360°. So $30n = 360$ and $n = 12$.

Check with the interior formula: a 12-sided polygon has angle sum $(12 - 2) \cdot 180° = 1800°$, and $1800° / 12 = 150°$.

### Two mistakes to avoid

**Common mistake.** **Taking the exterior angle to be 360° minus the interior angle.** The word "exterior" suggests everything outside the corner, which for a 60° corner would be 300°. But the exterior angle is measured from the *extension* of one side to the other side, so it is $180° - 60° = 120°$. That is the turn a walker makes at the corner. A turn of 300° would leave them facing back into the shape.

**Common mistake.** **Counting $n \cdot 180°$ for a polygon.** A tempting way to cut up a polygon is to pick a point $O$ in the middle and join it to all $n$ corners. That makes $n$ triangles, which suggests an angle sum of $n \cdot 180°$. But the angles of those triangles at $O$ are not angles of the polygon. They go once around $O$, total 360°, and must be subtracted: $n \cdot 180° - 360° = (n - 2) \cdot 180°$.

## Where the theorem fails: triangles on a sphere

To see that the parallel axiom was doing real work in the proof, we can look at a surface where the axiom is false and ask whether the theorem survives.

On a sphere, the straightest possible paths are the **great circles**: circles whose center is the center of the sphere, such as the equator or a line of longitude on a globe. They play the role of straight lines, and a triangle on a sphere is made of three arcs of great circles.

[Figure 8, image: A triangle on a sphere whose sides are a quarter of the equator and two lines of longitude from the north pole. Every corner is a right angle, so the angles add to 270°.]

Start at the north pole and walk straight down to the equator. Turn left through a right angle and walk a quarter of the way around the world along the equator. Turn left through a right angle again and walk straight back to the pole. You arrive at a right angle to the direction you left in, because the two lines of longitude are a quarter turn apart. The triangle you have walked has three right angles, and its angle sum is 270°.

The proof breaks at its first sentence: let $\ell$ be the line through $C$ parallel to $AB$. Take $C$ to be the north pole and $AB$ the equator. Every great circle through the pole is a line of longitude, and every line of longitude crosses the equator, so no line through $C$ avoids $AB$. Any two great circles meet, so a sphere has no parallel lines at all, and the parallel axiom, which promises exactly one, is false there.

The sphere breaks some of Euclid's other axioms too (two great circles meet twice), which is why property 3 of the half turn fails on it. The hyperbolic plane keeps the other axioms and has many parallels through each point, and there every triangle has an angle sum below 180°. An angle sum of exactly 180° goes with exactly one parallel.

The amount by which a spherical triangle exceeds 180° follows a precise law, published by Albert Girard in 1629, which I'll state without proof. On a sphere of radius $R$, a triangle of area $S$ has angle sum

$$
180° + \frac{180°}{\pi} \cdot \frac{S}{R^2}.
$$

Our three-right-angle triangle covers one eighth of the sphere. The whole sphere has area $4\pi R^2$, so $S = \pi R^2 / 2$, and the formula gives $180° + \frac{180°}{\pi} \cdot \frac{\pi}{2} = 270°$, as we found by walking.

**Pause and try.** The Earth is close to a sphere of radius 6371 km. Surveyors lay out a triangle on its surface with an area of 10,000 km², which means sides of roughly 150 km. By how much does its angle sum exceed 180°? Would the paper-tearing experiment detect an excess of that size?

**Answer.** Here $S / R^2 = 10{,}000 / 6371^2 \approx 0.000246$. Multiplying by $180° / \pi \approx 57.3°$ gives an excess of about $0.014°$. Torn paper cannot show a seventieth of a degree. This is the worry from the start of the lesson made real: on a slightly curved surface, careful measurement of small triangles would report 180° and be wrong. Only an argument can tell the two cases apart, and it has to use something that separates flat from curved, which is the parallel axiom.

## Looking back

The hard part of this lesson was the move from "the corners seem to fit" to "the corners must fit". What made it possible was adding one new line to the picture, the parallel through the top corner, chosen so that each base angle has an equal partner on it. Adding a line that the problem did not give you is one of the most common devices in geometry, and it is usually found the way we found it here, by doing the experiment carefully and asking where the pieces end up.

In the next lesson, on [the Pythagorean theorem](@geometry/pythagorean-theorem), this theorem supplies one necessary step: in a right triangle the two angles other than the right angle add to 90°.

**Check yourself.** A triangle has an exterior angle of 130° at one corner. One of the interior angles at the *other* two corners is 55°. What are the three interior angles?
Answer: 50°, 55° and 75° The interior angle beside the exterior one is $180° - 130° = 50°$. The exterior angle equals the sum of the other two interior angles, so the last one is $130° - 55° = 75°$. Check: $50 + 55 + 75 = 180$.

**Recap.** - The interior angles of every triangle in the plane add to exactly 180°.
- The proof draws the line through one corner parallel to the opposite side. Alternate angles put copies of the other two angles on it, and the three fill a straight angle.
- An exterior angle of a triangle equals the sum of the two interior angles at the other corners.
- A convex polygon with $n$ sides has interior angles adding to $(n - 2) \cdot 180°$ and exterior angles adding to 360°.
- All of this depends on the parallel axiom. On a sphere there are no parallel lines, and the angles of a triangle add to more than 180°.

## Problems

**Problem 1.** A triangle has angles of 48° and 67°. Find the third angle.

**Solution.** The angles add to 180°, so the third is $180° - 48° - 67° = 65°$. Check: $48 + 67 + 65 = 180$.

**Problem 2.** The angles of a triangle are $x$, $x + 20$ and $2x$ degrees. Find them.

**Solution.** Adding the three gives $x + (x + 20) + 2x = 4x + 20$, and this must equal 180. So $4x = 160$ and $x = 40$. The angles are 40°, 60° and 80°. Check: $40 + 60 + 80 = 180$, and all are positive.

**Problem 3.** In a right triangle, one of the two angles other than the right angle is four times the other. Find both. Then explain why, in every right triangle, those two angles add to 90°.

**Solution.** The three angles add to 180° and one of them is 90°, so the other two add to $180° - 90° = 90°$. That holds for every right triangle. Here the two are $y$ and $4y$, so $5y = 90$ and $y = 18$. The angles are 18° and 72°. Check: $90 + 18 + 72 = 180$.

**Problem 4.** The interior angles of a convex polygon add to 1980°. How many sides does it have?

**Solution.** We need $(n - 2) \cdot 180 = 1980$, so $n - 2 = 1980 / 180 = 11$ and $n = 13$.

[Figure 9, image: Two parallel lines joined by a path with one bend. The path leaves the top line at 35° and the bottom line at 45°. The angle at the bend is to be found.]

**Problem 5.** In the figure above, the lines $\ell$ and $m$ are parallel. The segment $TX$ makes an angle of 35° with $\ell$, and the segment $UX$ makes an angle of 45° with $m$. Find the angle $\angle TXU$ at the bend. (Hint: the problem did not give you enough lines.)

**Solution.** Draw the line $n$ through $X$ parallel to $\ell$. It is also parallel to $m$: if $n$ met $m$ at some point, then through that point there would be two different lines parallel to $\ell$, which the parallel axiom forbids. The line $n$ runs through the bend and splits $\angle TXU$ into an upper part and a lower part.

The line $TX$ crosses the parallel lines $\ell$ and $n$, and the 35° angle at $T$ and the upper part of the angle at $X$ are alternate angles: they lie between $\ell$ and $n$ on opposite sides of $TX$. So the upper part is 35°. In the same way the line $UX$ crosses the parallel lines $m$ and $n$, and the lower part is alternate to the 45° angle at $U$, so it is 45°. Therefore $\angle TXU = 35° + 45° = 80°$.

[Figure 10, image: A five-pointed star drawn with five straight lines through the points A, B, C, D, E. The line BE crosses AC at P and AD at Q. The five shaded angles at the tips are to be added.]

**Problem 6.** The five-pointed star above is made by joining $A$ to $C$, $C$ to $E$, $E$ to $B$, $B$ to $D$ and $D$ to $A$. It is not assumed to be symmetric. Prove that the five angles at its tips add to 180°. (Hint: look at the small triangle $APQ$, and use the exterior angle theorem twice.)

**Solution.** The triangle $APQ$ has the tip angle $\angle A$ at $A$, so by the angle sum theorem

$$
\angle A + \angle APQ + \angle AQP = 180°.
$$

We will show that $\angle APQ = \angle C + \angle E$ and $\angle AQP = \angle B + \angle D$.

Look at the triangle $PCE$. Its sides at $C$ run along $CA$ and $CE$, and its sides at $E$ run along $EB$ and $EC$, because $P$ lies on both $CA$ and $EB$. So its angles at $C$ and $E$ are the tip angles $\angle C$ and $\angle E$. Now $P$ lies on the segment $AC$ between $A$ and $C$, so the ray $PA$ is the extension of the side $CP$ past $P$. The angle between that extension and the side $PE$ is an exterior angle of triangle $PCE$ at $P$. Since $Q$ lies on the ray $PE$, that exterior angle is $\angle APQ$. By the exterior angle theorem, $\angle APQ = \angle C + \angle E$.

The same argument on the triangle $QBD$, whose angles at $B$ and $D$ are the tip angles there and whose side $DQ$ extends past $Q$ to $A$, gives $\angle AQP = \angle B + \angle D$.

Substituting, $\angle A + (\angle C + \angle E) + (\angle B + \angle D) = 180°$.

**Problem 7.** On a globe, a triangle has one corner at the north pole and two corners on the equator, 120° of longitude apart. Its sides are two lines of longitude and an arc of the equator. (a) Find the sum of its angles. (b) Name the first sentence of the proof of the triangle theorem that cannot be carried out for this triangle.

**Solution.** (a) Each line of longitude meets the equator at a right angle, so the two angles on the equator are 90° each. The angle at the pole is the angle between the two lines of longitude, which is 120°. The sum is $90° + 90° + 120° = 300°$.

(b) "Let $\ell$ be the line through $C$ parallel to the line $AB$." Every great circle through the pole $C$ is a line of longitude and crosses the equator, so no such $\ell$ exists.