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But the three planes are equal to the three opposite; therefore the three solids LQ, KR, AU are equal to one another.

For the same reason the three solids ED, DM, MT are also equal to one another.

Therefore, whatever multiple the base LF is of the base AF, the same multiple also is the solid LU of the solid AU.

For the same reason, whatever multiple the base NF is of the base FH, the same multiple also is the solid NU of the solid HU.

And, if the base LF is equal to the base NF, the solid LU is also equal to the solid NU; if the base LF exceeds the base NF, the solid LU also exceeds the solid NU; and, if one falls short, the other falls short.

Therefore, there being four magnitudes, the two bases AF, FH, and the two solids AU, UH, equimultiples have been taken of the base AF and the solid AU, namely the base LF and the solid LU, and equimultiples of the base HF and the solid HU, namely the base NF and the solid NU, and it has been proved that, if the base LF exceeds the base FN, the solid LU also exceeds the solid NU, if the bases are equal, the solids are equal, and if the base falls short, the solid falls short.

Therefore, as the base AF is to the base FH, so is the solid AU to the solid UH. [V. Def. 5] Q. E. D.


PROPOSITION 26.

On a given straight line, and at a given point on it, to construct a solid angle equal to a given solid angle.

Let AB be the given straight line, A the given point on it, and the angle at D, contained by the angles EDC, EDF, FDC, the given solid angle; thus it is required to construct on the straight line AB, and at the point A on it, a solid angle equal to the solid angle at D.

For let a point F be taken at random on DF, let FG be drawn from F perpendicular to the plane through ED, DC, and let it meet the plane at G, [XI. 11] let DG be joined, let there be constructed on the straight line AB and at the point A on it the angle BAL equal to the angle EDC, and the angle BAK equal to the angle EDG, [I. 23] let AK be made equal to DG,

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