Mathematical Collections (Commandino 1588)

we have demonstrated that, as AB is to BG, so is AC to CG: but as AB is to BG, so is KA to MG: as AC is to CG so is AL to NG. Because of the similarity of the triangles ALC and CNG, it will be as KA is to MG, so is AL to NG; and by permuting, as KA is to AL, so is MG to GN. But KA is equal to AL. Therefore, MG will also be equal to GN. But it is also unequal, which cannot be. Therefore, the straight line LC meets FG at M. Because KB and LC meet each other at FG at point M, the perpendicular drawn from it will fall to the same point G. And everything will happen similarly as in the above.

B This theorem is useful for the harmonic mean] He used a theorem improperly for a problem.

PROBLEM VI. PROPOSITION X.

A Given straight lines AB and BG, to find the mean in harmonic mean.

A geometric diagram for Proposition X shows a triangle with vertices A, D, B. A line segment extends from A to B, with points C and G marked on it. A perpendicular line segment extends from G to D. Lines connect D to A and D to B. A line segment FCE intersects the triangle, with F on line DB and E on a line extending from A.

But if the extremes AB and BG are given, and we seek the mean, by joining BD, and from point G drawing GF at right angles to GB, and from point F drawing FCE to E, we will have CB as the mean between AB and BG. And the demonstration is manifest.

PROBLEM VII. PROPOSITION XI.

A Given straight lines CB and BE, to find the major extreme in harmonic mean.

A geometric diagram for Proposition XI. A vertical line segment extends from D to E. A horizontal line segment extends from C through E to B. A line connects D to B. A line connects C to a point D. A point F is marked on the line DB. A line connects C to F and extends to a point H. A small decorative figure is drawn at the bottom right of the diagram.

Given the straight lines CB and BE, we will find the major extreme by drawing DE and EF from point E at right angles, and joining BF and DC, and producing them to G. For the perpendicular GH, which is drawn from point G to the extension of BC, cuts off HB equal to that which we were seeking; for CD and BF will meet each other at the parts of G. For it is necessary to assume BE is greater than EC.