# Proof: By Euclid

• Let $ABC$, $DEF$, and $GHK$ be three rectilinear angles, of which the sum of any) two is greater than the remaining (one), (the angles) being taken up in any (possible way) - (that is), $ABC$ and $DEF$ (greater) than $GHK$, $DEF$ and $GHK$ (greater) than $ABC$, and, further, $GHK$ and $ABC$ (greater) than $DEF$.
• And let $AB$, $BC$, $DE$, $EF$, $GH$, and $HK$ be equal straight lines.
• And let $AC$, $DF$, and $GK$ have been joined.
• I say that it is possible to construct a triangle out of (straight lines) equal to $AC$, $DF$, and $GK$ - that is to say, that (the sum of) any two of $AC$, $DF$, and $GK$ is greater than the remaining (one).

• Now, if the angles $ABC$, $DEF$, and $GHK$ are equal to one another then (it is) clear that, (with) $AC$, $DF$, and $GK$ also becoming equal, it is possible to construct a triangle from (straight lines) equal to $AC$, $DF$, and $GK$.
• And if not, let them be unequal, and let $KHL$, equal to angle $ABC$, have been constructed on the straight line $HK$, at the point $H$ on it.
• And let $HL$ be made equal to one of $AB$, $BC$, $DE$, $EF$, $GH$, and $HK$.
• And let $KL$ and $GL$ have been joined.
• And since the two (straight lines) $AB$ and $BC$ are equal to the two (straight lines) $KH$ and $HL$ (respectively), and the angle at $B$ (is) equal to $KHL$, the base $AC$ is thus equal to the base $KL$ [Prop. 1.4].
• And since (the sum of) $ABC$ and $GHK$ is greater than $DEF$, and $ABC$ equal to $KHL$, $GHL$ is thus greater than $DEF$.
• And since the two (straight lines) $GH$ and $HL$ are equal to the two (straight lines) $DE$ and $EF$ (respectively), and angle $GHL$ (is) greater than $DEF$, the base $GL$ is thus greater than the base $DF$ [Prop. 1.24].
• But, (the sum of) $GK$ and $KL$ is greater than $GL$ [Prop. 1.20].
• Thus, (the sum of) $GK$ and $KL$ is much greater than $DF$.
• And $KL$ (is) equal to $AC$.
• Thus, (the sum of) $AC$ and $GK$ is greater than the remaining (straight line) $DF$.
• So, similarly, we can show that (the sum of) $AC$ and $DF$ is greater than $GK$, and, further, that (the sum of) $DF$ and $GK$ is greater than $AC$.
• Thus, it is possible to construct a triangle from (straight lines) equal to $AC$, $DF$, and $GK$.
• (Which is) the very thing it was required to show.

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:
@Fitzpatrick