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. 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.
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. 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.
Euclid. Euclid's Elements. Sir Thomas Little Heath. New York. Dover. 1956.
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