|To describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid on a given straight line.|
Let AB be the given straight line and CD the given parallelepipedal solid.
It is required to describe on the given straight line AB a parallelepipedal solid similar and similarly situated to the given parallelepipedal solid CD.
|Construct the solid angle contained by the angles BAH, HAK, and KAB at the point A on the straight line AB equal to the solid angle C so that the angle BAH equals the angle ECF, the angle BAK equals the angle ECG, and the angle KAH equals the angle GCF,so that EC is to CG as BA is to AK, and GC is to CF as KA is to AH.||XI.26|
|Therefore, ex aequali, EC is to CF as BA is to AH.||V.22|
|Complete the parallelogram HB and the solid AL.|
|Now since EC is to CG as BA is to AK, and the sides about the equal angles ECG and BAK are thus proportional, therefore the parallelogram GE is similar to the parallelogram KB. For the same reason the parallelogram KH is similar to the parallelogram GF, and also FE is similar to HB.|
|Therefore three parallelograms of the solid CD are similar to three parallelograms of the solid AL. But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms, therefore the whole solid CD is similar to the whole solid AL.||XI.Def.9|
|Therefore on the given straight line AB there has been described AL similar and similarly situated to the given parallelepipedal solid CD.|
|Q. E. F.|
Next proposition: XI.28