If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles.
若有两个相等的平面角，在其顶点处分别设立两条升高直线，使得它们与原直线所成角对应相等，在升高直线上任取两点，从这两点向原角所在平面作垂线，再从垂足向原角顶点连线，则这些连线与升高直线所成角相等。
Let the angles BAC, EDF be two equal rectilineal angles, and from the points A, D let the elevated straight lines AG, DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle MDE to the angle GAB and the angle MDF to the angle GAC, let points G, M be taken at random on AG, DM, let GL, MN be drawn from the points G, M perpendicular to the planes through BA, AC and ED, DF, and let them meet the planes at L, N, and let LA, ND be joined; I say that the angle GAL is equal to the angle MDN. 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.
作AH等于DM，过H作HK平行于GL，则HK垂直于平面BAC。
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. Similarly we can prove that AB is also equal to DE.
从K、N向AC、DF、AB、DE作垂线，连接HC、CB、MF、FE。由勾股定理得角HCA和角DFM均为直角。
Since then AC is equal to DF, and AB to DE, the two sides CA, AB are equal to the two sides FD, DE. But the angle CAB is also equal to the angle FDE; therefore the base BC is equal to the base EF, the triangle to the triangle, and the remaining angles to the remaining angles; [I. 4] therefore the angle ACB is equal to the angle DFE. But the right angle ACK is also equal to the right angle DFN; therefore the remaining angle BCK is also equal to the remaining angle EFN. For the same reason the angle CBK is also equal to the angle FEN. Therefore BCK, EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equal to EF; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore CK is equal to FN. But AC is also equal to DF; therefore the two sides AC, CK are equal to the two sides DF, FN; and they contain right angles.
三角形HAC与MDF有两角及一边相等，故AC等于DF，同理AB等于DE。进而三角形ABC与DEF全等，得BC等于EF，角ACB等于角DFE。
Therefore the base AK is equal to the base DN. [I. 4] And, since AH is equal to DM, the square on AH is also equal to the square on DM. But the squares on AK, KH are equal to the square on AH, for the angle AKH is right; [I. 47] and the squares on DN, NM are equal to the square on DM, for the angle DNM is right; [I. 47] therefore the squares on AK, KH are equal to the squares on DN, NM; and of these the square on AK is equal to the square on DN; therefore the remaining square on KH is equal to the square on NM; therefore HK is equal to MN. And, since the two sides HA, AK are equal to the two sides MD, DN respectively, and the base HK was proved equal to the base MN, therefore the angle HAK is equal to the angle MDN. [I. 8] Therefore etc. PORISM.
由角ACK和角DFN均为直角，得角BCK等于角EFN，同理角CBK等于角FEN，故三角形BCK与EFN全等，得CK等于FN。再由AC等于DF，得AK等于DN。最后利用勾股定理和三角形HAK与MDN全等，证得角GAL等于角MDN。