Extremities of Line Segments containing three Plane Angles any Two of which are Greater than Other form Triangle

Proof

 * Euclid-XI-22a.png

Let $\angle ABC, \angle DEF, \angle GHK$ be plane angles such that the sum of any two is greater than the remaining one.

That is:
 * $\angle ABC + \angle DEF > \angle GHK$
 * $\angle DEF + \angle GHK > \angle ABC$
 * $\angle GHK + \angle ABC > \angle DEF$

Let the straight lines $AB, BC, DE, EF, GH, GK$ be equal.

Let $AC$, $DF$ and $GK$ be joined.

It is to be demonstrated that it is possible to construct a triangle from straight lines equal to $AC$, $DF$ and $HK$.

If $\angle ABC, \angle DEF, \angle GHK$ are equal to one another, then it is possible to construct an equilateral triangle from $AC$, $DF$ and $HK$.


 * Euclid-XI-22b.png

Otherwise, let $\angle ABC, \angle DEF, \angle GHK$ be unequal.

On the straight line $HK$ at the point $H$, let:
 * $\angle KHL$ be constructed equal to $\angle ABC$
 * $HL$ be constructed equal to one of $AB, BC, DE, EF, GH, GK$
 * $KL$ be joined.

We have that:
 * $AB$ and $BC$ are equal to $KH$ and $HL$
 * $\angle ABC = \angle KHL$

Therefore from :
 * $AC = KL$

We have that:
 * $\angle GHK + \angle ABC > \angle DEF$

while:
 * $\angle ABC = \angle KHL$

Therefore:
 * $\angle GHL > \angle DEF$

We have that:
 * $GH$ and $HL$ are equal to $DE$ and $EF$
 * $\angle GHL > \angle DEF$

Therefore from :
 * $GL > DF$

But:
 * $GK + KL > GL$

Therefore:
 * $GK + KL > DF$

But:
 * $KL = AC$

Therefore:
 * $AC + GK > DF$

Similarly it can be proved that:
 * $AC + DF > GK$

and that:
 * $DF + GK > AC$

Hence the result.