Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

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Again, let sides subtending equal angles be equal, as AB to DE;

I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and
further the remaining angle BAC is equal to the remaining angle EDF.

For, if BC is unequal to EF, one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles;

therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF,
and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle BHA is equal to the angle EFD.

But the angle EFD is equal to the angle BCA; therefore, in the triangle AHC, the exterior angle BHA is
equal to the interior and opposite angle BCA:

which is impossible. [I. 16]

Therefore BC is not unequal to EF,

and is therefore equal to it.

But AB is also equal to DE;
therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles;

therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angle
EDF. [I. 4]

Therefore etc.

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Q. E. D.

^{1 2}

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Proposition 27.

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If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

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For let the straight line EF falling on the two straight
lines AB, CD make the alternate angles AEF, EFD equal to one another;

I say that AB is parallel to CD.

For, if not, AB, CD when produced will meet either in the direction
of B, D or towards A, C.

Let them be produced and meet, in the direction of B, D, at G.

Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite
angle EFG: which is impossible. [I. 16]

Therefore AB, CD when produced will not meet in the direction of B, D.

Similarly it can be proved that neither will they meet
towards A, C.