(Edited by Dionysius Lardner, 1855)

Proposition XXII. Problem.

[Euclid, ed. Lardner, 1855, on `Google Books`]

(101) | Given three right lines (A, B and C) the sum of any two of which is greater than the third, to construct a triangle whose sides shall be respectively equal to the given lines. |

From any point D draw the right line D E equal to one of the given lines A (II), and from the same point draw D G equal to another of the given lines B, and from the point E draw E F equal to C. From the centre D with the radius D G describe a circle, and from the centre E with the radius E F describe another circle, and from a point K of intersection of these circles draw K D and K E.

It is evident, that the sides D E, D K and K E of the triangle D K E are equal to the given right lines A, B and C.

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

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^{★}★^{★}In this solution Euclid assumes that the two circles will have at least one point of intersection. To prove this, it is only necessary to show that a part of one of the circles will be within, and another part without the other (58).Since D E and E K or E L are together greater than D K, it follows, that D L is greater than the radius of the circle K G, and therefore the point L is outside the circle. Also, since D K and E K are together greater than D E, if the equals E K and E H be taken from both, D H is less than D K, that is, D H is less than the radius of the circle, and therefore the point H is within it. Since the point H is within the circle and L without it, the one circle must intersect the other.

It is evident, that if the sum of the lines B and C were equal to the line A, the points H and K would coincide; for then the sum of D K and K E would equal D E. Also, if the sum of A and C were equal to B, the points K and L would coincide; for then D K would be equal to E K and D E, or to L D. It will hereafter appear, that in the former case the circles would touch externally, and in the latter internally.

If the line A were greater than the sum of B and C, it is easy to perceive that the circle would not meet, one being wholly outside the other; and if B were greater than the sum of A and C, they would not meet, one being wholly within the other.

If the three right lines A B C be equal, this proposition becomes equivalent to the first, and the solution will be found to agree exactly with that of the first.