# Ask Uncle Colin: Bridges, Donkeys and Triangles

Dear Uncle Colin,

I’m struggling to understand why, if you know a triangle has two sides the same, the base angles must be the same. Can you explain?

-- I’m Struggling Over Some Coherent Explanation Leveraging Equal Sides

Hi, ISOSCELES, and thanks for your message!

There are several good proofs of this, but my favourite is known as the Bridge of Asses – reputedly so-called because it’s the first proof in Euclid that isn’t obvious; the implication, rather un-FCM-ly, is that if you can’t get past this proof, you’re a bit of a donkey.

I don’t subscribe to that point of view, obviously.

So, the proof is based on the idea that two triangles are congruent - identical - if every side of one of them has the same length as a corresponding side on the other. The *nugget* of the proof is to show that an isosceles triangle is congruent to its mirror image - which means the angles at the base have to be the same as each other.

Let’s go through it in a bit more detail: suppose your isosceles triangle is ABC, and sides AB and AC are the same length ((It really doesn’t matter how we label things.)). We’re going to think about triangle ABC and its mirror image, ACB.

The first side in the first triangle is AB, which is the same length as the first side in the second triangle, AC, because those are the sides we picked to be the same length.

The second side in the first triangle is BC, which is the same length as the second side in the second triangle, CB, because they’re the same side!

The final side in the first triangle is CA, which is the same length as the final side in the second triangle, BA - again, because we picked those to be the same length.

So! All three sides in either triangle correspond to a side in the other, which means they’re congruent. It’s not just the sides that correspond, though, the angles do, too!

So the first angle in the first triangle, CAB, is equal to the first angle in the second, BAC. That’s not a surprise: they’re both the apex of the triangle.

More interestingly, the second angle in the first triangle, ABC, is equal to the second angle in the second triangle, ACB. *And that’s what we were trying to prove all along*! So the proof is done.

We don’t need to write all of that in the proof, though. Instead, you’d say something like:

AB = AC (given)

BC = CB (identity)

CA = BA (given)

So ABC is congruent to ACB (side-side-side).

Therefore, $A\hat BC$ = $A\hat CB$. QED.

You can see why it’s considered a tough proof! All of those paragraphs condensed down into five lines - but if you read it carefully, you can see each of the lines means the same thing as one of the paragraphs above.

I hope that helps!

-- Uncle Colin

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