PROP. VIII. PROBLEM.

To find a third proportional to two given right lines.

A

Let A, B be two given right lines; it is required to find a third proportional to them.

Draw the two indefinite right lines CF, CG, making any angle C, and take CD equal to A, and CE, DF each equal B (I. 3.) :

Alfo join the points D, E, and make FG parallel to PE (I. 27.); and EG will be the third proportional required.

For, fince CFG is a triangle, and DE is parallel to FG (by Conft.) CD will be to DF as CE is to EG (VI. 3.)

But DF is equal to CE, by conftruction; therefore CD is to CE as CE is to EG.

And, fince CD is equal to A, and CE to B (by Conft.). A will be to B as B is to EG.

Q. E. D.

SCHOLIUM. A third proportional to two given right lines, may also be found by means of the laft propofition.

For let AD, DC be two given right lines, one of which DC is perpendicular to the other.

Then

Then if CB be drawn at right angles to AC, and AD be produced till it meets the former in B, DB will be a third proportional to AD and DC as was required.

Again, let BD, DC be the two given right lines, placed at right angles to each other, as before:

Then, if CA be drawn at right angles to BC, and BD be produced till it meets the former in A, DA will be a third proportional to BD and DC.

To find a fourth proportional to three given right lines.

Let A, B, C be three given right lines; it is required to find a fourth proportional to them.

Draw the two indefinite right lines DG, DH, making any angle D; and take DE equal to A, DF to B, and EG to c (I. 3.)

Join the points E, F, and make GH parallel to E F (1.27.); and FH will be the fourth proportional required.

. For, fince DGH is a triangle, and EF is parallel to GH (by Conft.) DE will be to DF as EG is to FH (VI. 3.)

But DE is equal to A, DF to B, and EG to C (by Const.); therefore A will be to в as C is to FH (V. 9.)

Q.E.D.

PRO P. X. PROBLEM.

To find a mean proportional between two given right lines.

E D

Let A, B be two given right lines; it is required to find a mean proportional between them.

Draw the indefinite right line CG, in which take CE equal to A, and ED equal to в (I. 3.)

Upon CD defcribe the femi-circle CFD, and from the point E erect the perpendicular EF (I. 11.); and it will be the mean proportional required.

For join the points CF, FD:

1

Then, because DFC is a femi-circle, the angle, CFD is a right angle (III. 16.), and confequently the triangle CDF is rectangular.

And fince a perpendicular from the right angle is a mean proportional between the segments of the hypote nufe (VI. 7.), EF will be a mean proportional to CE,

ED.

But CE is equal to A, and ED to B (by Conft.); therefore EF will be a mean proportional to A and B, as was to be fhewn.

SCHOLIUM. If CD, DE be the two given lines, DF will be a mean proportional to them.

And if CD, CE be the given lines, CF will be a mean proportional to them (VI. 7.)

PROP.

PROP, XI. PROBLEM.

Ta divide a given right line into two parts which fhall have the fame ratio with two given right lines.

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Let CD be a given right line, and A, B two other given right lines; it is required to divide CD into two parts which shall be to each other in the ratio of A to B.

Draw the indefinite right line CG, making any angle with CD; and make CF equal to A, and FG to B (I. 3.) Join the points GD; and make FE parallel to GD (I.27.); and it will divide CD in the point E as was required.

For, fince CDG is a triangle, and FE is parallel to GD (by Conft.), CE will be to ED as CF is to FG (VI. 3.) But CF is equal to A, and FG to B (by Conft.); there fore CE will be to ED as A is to B (V. 9.)

Q. E. D. SCHOLIUM. In nearly the fame manner may a third, fourth, or any other part be cut off from a given line CD.

.

For if CG be made the fame multiple of CF that CD is of the part, and GD, FE be drawn as above, CE will be the part required.

PRO P. XII. THEOREM.

If four right lines be proportional, the rectangle of the extremes will be equal to the rectangle of the means; and if the rect angle of the extremes be equal to the rectangle of the means, the four lines will be proportional.

Let AB be to CD as E is to F; then will the rectangle of AB and F, be equal to the rectangle of CD and E.

For make вH perpendicular to AB, and equal to r (I. 10. 3.), and DG perpendicular to CD, and equal to E; and in DG take DK equal to BH (I. 3.) and draw KL parallel to CD (I. 27.)

Then, fince parallelograms of equal altitudes, are to each other as their bafes (VI. 1.) the parallelogram AH will be to the parallelogram CK, as AB is to CD.

And fince AB is to CD as E is to F (by Hyp.), or as DG is to DK (by Conft.), the parallelogram AH will also be to the parallelogram CK as DG is to DK (V. 11.)

But parallelograms of the fame base, are to each other as their altitudes (VI. 2.); whence the parallelogram CG will also be to the parallelogram CK as DG is to DK.

And