And because ratios which are the fame to the fame ratio are the fame to each other (V. 11.), the parallelogram AH will be to the parallelogram CK, as the parallelogram CG is to the parallelogram CK.

But magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.); whence the rectangle AH is equal to the rectangle CG.

Again, let AH, the rectangle of the extremes, be equal to CG, the rectangle of the means; then will AB be to CD

as E is to F.

For, let the fame construction be made as before :

Then, fince rectangles of equal altitudes are to each other as their bases (VI. 2.), the rectangle AH will be to the rectangle CK as AB is to CD.

And, because the rectangle AH is equal to the rectangle CG (by Hyp.), CG will also be to CK as AB is to CD (V.9.) But CG is to CK as DG is to DK or BH (VI. 2.) confequently AB will be to CD as DG is to BH (V. 11.)

And fince DG is equal to E, and BH to F, AB will be to CD as E is to F (V. 9.)

Q. E. D.

1

PRO P. XIII. THEOREM.

If three right lines be proportional, the rectangle of the extremes will be equal to the fquare of the mean and if the rectangle of the extremes be equal to the fquare of the mean, the three lines will be proportional.

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

For make BK perpendicular to AB (I. 10.) and equal to E (I. 3.); and upon CD describe the fquare CG (II. 1.), and make F equal to CD.

Then, fince CD is equal to F, and AB is to CD as CD is to E (by Hyp.), AB will also be to CD as F is to E (V.9.) And, fince these four lines are proportional, the rectangle of AB and E will be equal to the rectangle of CD and F (VI. 12.)

But the rectangle of CD and F is equal to the fquare of CD, because CD is equal to F; therefore, also, the rectangle of AB and E will be equal to the fquare of CD.

Again, if the rectangle of AB and E be equal to the fquare of CD; AB will be to CD as CD is to E.

For let the fame conftruction be made as before:

Then, fince the rectangle of AB and E is equal to the fquare of CD (by Hyp.), and the fquare of CD is equal to the rectangle of CD and F (II. 2.), the rectangle of AB and E will also be equal to the rectangle of CD and F.

But if the rectangle of the extremes be equal to that of the means, the four lines are proportional (VI. 12.); . whence AB is to CD as F is to E.

And fince CD is equal to F, by conftruction, AB will be to CD as CD is to E.

Q; E. D.

PROP. XIV. THEOREM.

Equal parallelograms and triangles have their fides about equal angles reciprocally proportional; and if the fides about equal angles are reciprocally proportional, the parallelograms and triangles will be equal.

Let AB, AC be two equal parallelograms, and DFA, AEG two equal triangles, having the angle DAF equal to the angle GAE; then will the fide DA be to the fide aɛ, as the fide AG is to the fide AF.

For let the fides DA, AE be placed in the fame right line, and complete the parallelogram AK.

Then, because the angles DAF, FAE are equal to two right angles (I. 13.), and the angle FAE is equal to the angle DAG (I. 15.), the angles DAF, DAG are alfo equal to two right angles; and confequently FAG is a right line.

And fince the parallelogram AB is equal to the parallelogram AC (by Hyp.), and AK is another parallelogram, AB is to AK as AC is to AK (V. 9.)

But AB is to AK as DA to AE (VI. 1.), and Ac to AK as AG to AF; whence DA is to AE as AG is to AF (V. 11.)

And, if FE be joined, it may be fhewn, in like manner, that the triangle DAF is to the triangle AFE as the triangle GAE is to the triangle AFE; and DA to AE as AG to AF.

Again, let the angle DAF be equal to the angle GAE, and the fide DA to the fide AE as the fide AG is to the fide AF; then will the parallelogram AB be equal to the parallelogram AC, and the triangle DFA to the triangle GAE.

For fince DA is to AE as AG to AF (by Hyp.), and DA to AE as AB to AK (VI. 1.), AG will be to AF as AB to AK (V. II.)

But AG is to AF as AC to AK (VI. 1.); whence Ab is to AK as AC to AK (V. 11.); and confequently the parallelogram AB is equal to the parallelogram AC (V. 10.)

And fince triangles are the halves of parallelograms, which have the fame base and altitude, the triangle DFA will be equal to the triangle GAE.

Q. E. D.

COROLL. The fides of equal rectangles are reciprocally proportional; and if the fides are reciprocally proportional, the rectangles will be equal.

PROP. XV. PROBLEM.

Upon a given right line to defcribe a rectilineal figure, fimilar, and fimilarly fituated, to a given rectilineal figure.

Let AB be the given right line, and CDEFG the given rectilineal figure; it is required to defcribe a rectilineal figure upon AB, which shall be fimilar, and fimilarly fituated to CDEFG.

Join DG, DF; and at the points A, B, make the angles BAL, ABL equal to the angles DCG, CDG (I. 20.)

In like manner, at the points B, L, make the angles BLK, LBK equal to the angles DGF, GDF; and BKH, KBH equal to DFE, FDE.

Then, because two angles in one triangle ar equal to two angles in another, each to each, the remaining angles in each of the correfponding triangles, will also be equal (I. 28. Cor.)

And fince the angles ALB, BLK, are equal to the angles CGD, DGF, and LKB, BKH to GFD, DFE, the angle ALK will be equal to the angle CGF, and the angle LKH to the angle GFE.

And, in the fame manner, it may be fhewn that the angles KHB, HBA and BAL are equal to the angles FED, EDC and DCG.