Let point A be the given point, and line BC the given straigt line.

It is required to place a straight line AL equal to the given straight line BC with one end at the point A.

Join the straight line AB from the point A to the point B, and construct the equilateral DAB.

Produce the straight line AE and straight line BF in a straight line with DA and DB.

Describe the circle CGH with center B, and again, descibe the circle GKL with center D.

Since the point B is the center of the circle CGH, therefore, line BC equals line BG.

Again, since the point D is the center of the circle GKL, therefore, line DL equals line DG.

And in these, DA equals DB, therefore, the remainder line AL equals the remainder line BG.

But line BC was also proved equal to line BG, therefore each of the straight lines line AL and line BC equals line BG.

And things which equal the same thing also equal one another, therefore, line AL also equals line BC.

Therefore, the straight line AL equal to the given straight line BC has been placed with one end at the given point A.

Dark
Light