Kitap 11
34
Now the solid AB is equal to the solid CD; therefore CM is also greater than AG.
34
Let then CT be made equal to AG, and let the parallelepipedal solid VC be completed on NQ as base and with CT as height.
34
Now, since the solid AB is equal to the solid CD, and CV is outside them, while equals have to the same the same ratio, [V. 7] therefore, as the solid AB is to the solid CV, so is the solid CD to the solid CV.
34
But, as the solid AB is to the solid CV, so is the base EH to the base NQ, for the solids AB, CV are of equal height; [XI. 32] and, as the solid CD is to the solid CV, so is the base MQ to the base TQ [XI. 25] and CM to CT [VI. 1]; therefore also, as the base EH is to the base NQ, so is MC to CT.
34
But CT is equal to AG; therefore also, as the base EH is to the base NQ, so is MC to AG.
34
Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.
34
Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD.
34
Let the sides which stand up be again at right angles to the bases.
34
Now, if the base EH is equal to the base NQ, and, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, therefore the height of the solid CD is also equal to the height of the solid AB.
