Let AH be made equal to DM, and let HK be drawn through the point H parallel to GL.
But GL is perpendicular to the plane through BA, AC;
therefore HK is also perpendicular to the plane through BA, AC. [XI. 8]
From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined.
Since the square on HA is equal to the squares on HK, KA, and the squares on KC, CA are equal to the square on KA, [I. 47]
therefore the square on HA is also equal to the squares on HK, KC, CA.
But the square on HC is equal to the squares on HK, KC; [I. 47]
therefore the square on HA is equal to the squares on HC, CA.
Therefore the angle HCA is right. [I. 48]
For the same reason the angle DFM is also right.
Therefore the angle ACH is equal to the angle DFM.
But the angle HAC is also equal to the angle MDF.
Therefore MDF, HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that subtending one of the equal angles, that is, HA equal to MD;
therefore they will also have the remaining sides equal to the remaining sides respectively. [I. 26]
Therefore AC is equal to DF.