Isosceles Triangle has Two Equal Angles

Theorem
In isosceles triangles, the angles at the base are equal to each other.

Also, if the equal straight lines are extended, the angles under the base will also be equal to each other.

Proof


Let $$\triangle ABC$$ be an isosceles triangle whose side $$AB$$ equals side $$AC$$.

We extend the straight lines $$AB$$ and $$AC$$ to $$D$$ and $$E$$ respectively.

Let $$F$$ be a point on $$BD$$.

We cut off from $AE$ a length $AG$ equal to $$AD$$.

We draw line segments $$FC$$ and $$GB$$.

Since $$AF = AG$$ and $$AB = AC$$, the two sides $$FA$$ and $$AC$$ are equal to $$GA$$ and $$AB$$ respectively.

They contain a common angle, that is, $$\angle FAG$$.

So by Triangle Side-Angle-Side Equality‎, $$\triangle AFC = \triangle AGB$$.

Thus $$FC = GB$$, $$\angle ACF = \angle ABG$$ and $$\angle AFC = \angle AGB$$.

Since $$AF = AG$$ and $$AB = AC$$, then $$BF = CG$$.

But $$FC = GB$$, so the two sides $$BF, FC$$ are equal to the two sides $$CG, GB$$ respectively.

Then $$\angle BFC = \angle CGB$$ while $$CB$$ is common to both.

Therefore by Triangle Side-Angle-Side Equality‎, $$\triangle BFC = \triangle CGB$$.

Therefore $$\angle FBC = \angle GCB$$ and $$\angle BCF = \angle CBG$$.

So since $$\angle ACF = \angle ABG$$, and in these $$\angle BCF = \angle CBG$$, then $$\angle ABC = \angle ACB$$.

But $$\angle ABC$$ and $$\angle ACB$$ are at the base of $$\triangle ABC$$.

Also, we have already proved that $$\angle FBC = \angle GCB$$, and these are the angles under the base of $$\triangle ABC$$.

Hence the result.

This theorem is the famous "pons asinorum", that is, the "bridge of donkeys", supposedly for two reasons:
 * 1) The diagram accompanying it is supposed to look a bit like a bridge;
 * 2) If you can't cross this bridge (i.e. understand this theorem), you're supposedly a bit of a donkey.

Don't blame them - this is one of the more tortuous ways of proving this. Commentators have speculated that Euclid didn't know what he was doing when he wrote this.