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

For the same reason HK is also parallel to AB.

Therefore HEBK is a parallelogram; therefore HK is equal to EB. [I. 34]

But EB is equal to EA; therefore AE is also equal to HK.

But AH is also equal to HD; therefore the two sides EA, AH are equal to the two sides KH, HD respectively, and the angle EAH is equal to the angle KHD; therefore the base EH is equal to the base KD. [I. 4]

Therefore the triangle AEH is equal and similar to the triangle HKD.

For the same reason the triangle AHG is also equal and similar to the triangle HLD.

Now, since two straight lines EH, HG meeting one another are parallel to two straight lines KD, DL meeting one another, and are not in the same plane, they will contain equal angles. [XI. 10]

Therefore the angle EHG is equal to the angle KDL.

And, since the two straight lines EH, HG are equal to the two KD, DL respectively, and the angle EHG is equal to the angle KDL, therefore the base EG is equal to the base KL; [I. 4] therefore the triangle EHG is equal and similar to the triangle KDL.

For the same reason the triangle AEG is also equal and similar to the triangle HKL.

Therefore the pyramid of which the triangle AEG is the base and the point H the vertex is equal and similar to the pyramid of which the triangle HKL is the base and the point D the vertex. [XI. Def. 10]

And, since HK has been drawn parallel to AB, one of the sides of the triangle ADB,