Book 11
34
But parallelepipedal solids on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AB is equal to the solid CD.
34
Next, let the base EH not be equal to the base NQ, but let EH be greater; therefore the height of the solid CD is also greater than the height of the solid AB, that is, CM is greater than AG.
34
Let CT be again made equal to AG, and let the solid CV be similarly completed.
34
Since, as the base EH is to the base NQ, so is MC to AG, while AG is equal to CT, therefore, as the base EH is to the base NQ, so is CM to CT.
34
But, as the base EH is to the base NQ, so is the solid AB to the solid CV, for the solids AB, CV are of equal height; [XI. 32] and, as CM is to CT, so is the base MQ to the base QT [VI. 1] and the solid CD to the solid CV. [XI. 25]
34
Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV; therefore each of the solids AB, CD has to CV the same ratio.
34
Therefore the solid AB is equal to the solid CD. [V. 9]
