A way to directly see that the interior angles of triangle sum to $180^\circ$?

Question

I'm looking for a way to look at a triangle, and perhaps visualize a few extra lines, and be able to see that the interior angles sum to $180^\circ$.

I can visualize that supplementary angles sum to $180^\circ$. I'd like to be able to see the interior angle sum similarly...

I can see that the exterior angles must sum to $360^\circ$, because if you walked around the perimeter, you would turn around exactly once (though I can tell this is true, I don't really see it). I also saw a proof on KA, where the exterior angles were superimposed, to show they summed to $360^{\circ}$ (though I'm not 100% comfortable with this one).

Finally, for $a$, $b$, and $c$ exterior angles $a+b+c=360$:

\begin{align} (180-a) + (180-b) + (180-c) & = 3\times 180 - (a+b+c) \\ & = 3\times 180 - 360 \\ & = 180 \\ \end{align}

But I find this algebra hard to see visually/geometrically. Is there proof that enables one to directly see that the interior angles of triangle sum to $180^\circ$?

A couple of secondary questions:

• am I visually deficient in my ability to imagine?
• or, am I asking too much of a proof, that I be able to see it, and that beimg able to tell that it is true should be enough...?

Answer 1

Since that fact about the angle sum is equivalent to the parallel postulate, any visualization is likely to include a pair of parallel lines. Here's one from wikipedia:

Answer 2

Draw a circle around a triangle. The central angle of each vertex is twice its size and the three central angles make the full round, that is $360$. Hence, the sum of interior angles of the triangle is one half of $360$, that is $180$.

Answer 3

Every triangle could be inscribed in a circle.

Each inscribed angle measures half the arc opposite to it.

The angle sum of the triangle inscribed in a circle is half the total arc measures around the circle.

The total arc measures around a circle is $360$ degrees.

Thus the angle sum of a triangle is $180$

Answer 4

Ethan Bolker's answer is the standard proof, but a more visual way to see the result is to tile the plane with copies of your triangle. It is so effective, I will not even include a drawing.

Just think about it... the tiling is made by three sets of parallel lines in the directions of the triangle sides, and they meet at vertices of the tiling, where you will then find two copies of each angle.

Edit. By popular request, there's a picture below:

Edit 2. I like Steven Gubkin's answer better :) That is what I used to call the "near-sighted method"

Answer 5

It isn't a proof, but if you fold the bottom two angles in, then fold the angle opposite the base down, it will just fit together with the other two to form a straight angle. Try it with a piece of paper. Rather nifty.

The proof I know and like is the one in Ethan Bolker's answer.

Answer 6

Imagine a triangle $ABC$ with two of the sides, say $AB$ and $BC$, rigid but meeting at the joint $B$, which is movable in the plane of the triangle, like a hinge sort of (also, you may fix one of these rigid sides, say $BC$, and rotate the other). Let the side $AC$ be made of a sufficiently elastic material (at least so that $AC\leq AB+BC$ ).

If you play with this imaginary object, you should soon see that as you increase $A \hat B C$, the other two angles eventually diminish (certainly for $A \hat B C>π/2$). Indeed, you may extend $\angle ABC$ as wide as you please close to $\pi$, so that the two rigid sides approach a single rectilinear segment; then the angle between them becomes arbitrarily close to $π$ and the other two angles are vanishing; in the limiting (degenerate) position the side $AC$ coincides with the segment formed by the rigid sides.

If you can already see that any point in this sequence of transformations of $\triangle ABC$ the sum of angles is conserved, then it is clear why for any triangle this sum is always $π$.

Answer 7

If you draw the triangle on a piece of paper, then cut off each corner, you can rearrange the corners so that the angles add up to a straight line.

Here is my simplest visual explanation courtesy of Paint:

Answer 8

This is very similar to Ethan Bolker's and Rodrigo A. Pérez's answers, but I made a small animation to illustrate a version that I like.

Answer 9

In a similar vein to your walking around the exterior and obtaining an external angle sum of 360 degrees, you can do the same by "walking" inside the triangle. The demonstration is best with physical barriers for sides.

Make a triangle using for example, blocks of wood or standing cardboard. Place a pen inside one corner against one side. Record the pen's orientation. Slide it along the side it is touching, and when you reach the end, swing it inside the triangle to rest against the other side. Repeat until the pen is back in the same position it started.

Observe that the pen has been spun exactly halfway around by spinning inside the angles of the triangle, so together the angles of the triangle must be 180 degrees.

This process can be extended to demonstrate physically the internal angle sum of any shape with two or more sides - for two sides the angle is zero as the pen cannot spin at all and for quadrilaterals or higher the angle is given by the usual formula, observed as one complete spin for every 360 degrees.

Answer 10

Not a proof, but a nice way to "see it" dynamically: extend the sides of the triangle so you can see their exterior angles.

Now "zoom out" really far away from the picture. The triangle will look like a point, and angles $1$, $2$, and $3$ will have to sum to $360$. However, since these together with the interior angles sum to $3\cdot 180$, the interior angles themselves must sum to $180$.

You could also think of the angles sliding along each other similar to how a camera shutter closes.

Answer 11

This explanation makes it clear why this particular theorem about triangles in the plane is deeply connected to the parallel postulate.

Consider two parallel lines. Lay two parallel transversals along them, so that the four lines determine a parallelogram. Then by the parallel postulate the two interior angles cut off on the same side by a transversal falling on two parallel lines are supplementary -- that is, they add up to two right angles. Consequently, the sum of angles in any parallelogram is four right angles, or $2π$.

Now it is clear that a parallelogram can be decomposed (along a diagonal) into two congruent triangles; and conversely, that two congruent triangles, plus a finite sequence of euclidean transformations, can be made to determine a parallelogram. In other words, two copies of a triangle determine a parallelogram. Consequently, the sum of angles in any triangle is half that of the related parallelogram, which is always $2π$. The result follows.

Edit. An even shorter way to see the connection to the parallel postulate is this. Let a straight line fall on two straight lines so that the two interior angles on one side of the transversal are less than supplementary; we have three lines here. Keep the transversal and either of the other two lines fixed, then rotate the last about its point of intersection with the transversal so that the two angles in question gradually approach supplementarity. Now, at all times when they're less than supplementary, we always have a triangle on that side (by the postulate). As you make the angles closer to supplementary, the distance from the transversal to the point where the lines meet increases indefinitely (and the angle at that meeting point becomes smaller and smaller). In the limit when the lines are parallel (i.e., they don't meet anywhere), the angles in question are just supplementary.

Answer 12

Euclid Elements, book I, proposition 32

I've reflected it, and used colours to try to to match the top answer here.

Though asymmetric, it's useful in proving other theorems, as it also shows: an exterior angle equals the sum of the opposite interior angles (blue and red).

Answer 13

If you accept that the sum $S$ of angles is the same in every triangle there’s another proof. Choose a point $D$ on $AB$. Now the sum of angles in the triangles $ADC$ and $DBC$ is $2S$:

$$2S=A+\angle ADC+\angle DCA+B+\angle DBC+\angle BCD=A+B+C+180,$$ hence $S=180$.