# Proof: By Euclid

• Let the (straight line) $AC$ have been cut in half at (point) $E$ [Prop. 1.10].
• And $BE$ being joined, let it have been produced in a straight line to (point) $F$.1 And let $EF$ be made equal to $BE$ [Prop. 1.3], and let $FC$ have been joined, and let $AC$ have been drawn through to (point) $G$.
• Therefore, since $AE$ is equal to $EC$, and $BE$ to $EF$, the two (straight lines) $AE$, $EB$ are equal to the two (straight lines) $CE$, $EF$, respectively.
• Also, angle $AEB$ is equal to angle $FEC$, for (they are) vertically opposite [Prop. 1.15].
• Thus, the base $AB$ is equal to the base $FC$, and the triangle $ABE$ is equal to the triangle $FEC$, and the remaining angles subtended by the equal sides are equal to the corresponding remaining angles [Prop. 1.4].
• Thus, $BAE$ is equal to $ECF$.
• But $ECD$ is greater than $ECF$.
• Thus, $ACD$ is greater than $BAE$.
• Similarly, by having cut $BC$ in half, it can be shown (that) $BCG$ - that is to say, $ACD$ - (is) also greater than $ABC$.
• Thus, for any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles.
• (Which is) the very thing it was required to show.

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### References

1. The implicit assumption that the point $F$ lies in the interior of the angle $ABC$ should be counted as an additional postulate (translator's note)