Let the three given straight lines be line AX, line BY, and line CZ, and let the sum of any two of these be greater than the remaining one, namely, line AX plus line BY greater than line CZ, line AX plus line CZ greater than line BY, and line BY plus line CZ greater than line AX.

It is required to construct triangle FKG out of straight lines equal to line AX, line BY, and line CZ.

Set out a straight line DE, terminated at point D but of infinite length in the direction of point E. Make line DF equal to line AX, line FG equal to line BY, and line GH equal to line CZ.

Describe the circle DKL with center F and radius DF. Again, describe the circle KLH with center G and radius GH. Join line KF and line KG.

I say that the triangle KFG has been constructed out of three straight lines equal to line AX, line BY, and line CZ.

Since the point F is the center of the circle DKL, therefore line DF equals line FK. But line DF equals line AX, therefore line FK also equals line AX.

Again, since the point G is the center of the circle KLH, therefore line GH equals line GK. But line GH equals line CZ, therefore line GK also equals line CZ.

And line FG also equals line BY, therefore the three straight lines line FK, line FG, and line GK equal the three straight lines line AX, line BY, and line CZ.

Therefore out of the three straight lines line FK, line FG, and line GK, which equal the three given straight lines line AX, line BY, and line CZ, the triangle FKG has been constructed.

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