For on the straight line AB and at the point A on it let the solid angle, contained by the angles BAH, HAK, KAB, be constructed equal to the solid angle at C, so that the angle BAH is equal to the angle ECF, the angle BAK equal to the angle ECG, and the angle KAH to the angle GCF;
and let it be contrived that, as EC is to CG, so is BA to AK, and, as GC is to CF, so is KA to AH. [VI. 12]
Therefore also, ex aequali, as EC is to CF, so is BA to AH. [V. 22]
Let the parallelogram HB and the solid AL be completed.