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AE, AF; whence AB is to AC as the rectangle of AD, AF is to the rectangle of AG, AE (V. 11 and 15.)

Again, let DFA, AEG be two triangles, having the angle DAF equal to the angle GAE, then will DFA be to AEG as the rectangle of DA, AF is to the rectangle of GA, AE.

For let the fides DA, AE be placed in the fame right line; and complete the parallelograms AB, AC, AK.

Then, as before, AB is to AC as the rectangle of DA, AF is to the rectangle of AG, AE.

But the triangles DFA, AEG are half the parallelograms AB, AC (I. 30.); whence DFA is to AEG as the rectangle of DA, AF is to the rectangle of GA, aɛ.

Q. E. D. SCHOLIUM. If the line EF be drawn, the latter part of this propofition may be proved from the triangles, independently of the former.

PROP. XIX. THEOREM.

The rectangles under the corresponding lines of two ranks of proportionals, are themfelves proportionals.

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Let AB be to CB as DE is to FF, and GH to GK as LM to Lt; then will the rectangle of AE, GH be to that of

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CB, GK as the rectangle of DE, LM is to that of FE, LN.

For draw BQ, ES at right angles to AB, DE (I. II.), and make BP equal to GH, BQ to GK, ER to LM, and Es to LN (I. 3.); and complete the parallelograms ap, CQ, DR and Fs.

Then fince parallelograms, of the fame altitude, are to each other as their bafès (VI. 1.), AP will be to CP as AB to CB; and DR to FR as DE to FE.

But AB is to CB as DE to FE (by Hyp.); whence AP will be to CP as DR to FR (V. 11.), or AP to DR às CP to FR (V. 15.)

And fince parallelograms of the fame base are to each other as their altitudes (VI. 2.), CP will be to co as BP to BQ; and FR to Fs as ER to Es.

Or, because BP, BQ are equal to GH, GK, and ER, ES to LM, LN (by Const.), CP will be to co as GH to GK; and FR to Fs as LM to LN (V. 9.)

But GH is to GK as LM to LN (by Hyp.), therefore cr will be to CQ as FR is to Fs (V. 11.), or CP to FR as CQ to FS (V. 15.)

And it has been before fhewn that AP is to DR as c to FR; whence AP is to DR as CQ to FS (V. 11.), or AP to CQ as DR to FS (V. 15.)

But AP is the rectangle of AB, GH; CQ of CB, GK; DR of DE, LM; and Fs of FE, LN (by Conft.); therefore the rectangle of AB, GH is to that of CB, GK as the rectangle of DE, LM is to that of FE, LN.

Q. E. D.

COROLL. The fquares of four proportional lines are themselves proportionals.

PROP.

PROP. XX. THEOREM.

The fides and diagonals of four proportional fquares, are themselves proportional.

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Let AC, EF, HL and QN be four proportional fquares; then will their fides and diagonals be alfo proportionals.

For, make s a fourth proportional to AB, EB and HK (VI. 9.); and draw the diagonals BD and KM.

Then, fince AB is to EB as HK is to s (by Conft.), AC will be to EF as HL is to the fquare of s (VI. 19. Cor.)

And because AC is to EF as HL is to QN (by Hyp.), HL will be to the fquare of s as HL is to QN, or the fquare of QK (V. 11.)

But magnitudes which have the fame ratio to the fame magnitude are equal to each other (V. 10.); whence the fquare of s is equal to the fquare of QK.

And fince equal fquares have equal fides (II. 3.), s is equal to QK; and confequently AB is to EB as HK to

to QK.

Again, because the triangles ABD, EBG are equiangu lar, AB will be to EB as BD to BG (VI. 5.)

And because the triangles HKM, QKP are alfo equiangular, HK will be to QK, as KM to KP (VI. 5.)

But AB has been fhewn to be to EB as HK is to OK; confequently BD is to BG as KM is to KP (V. 11.)

Q. E. D.

PROP. XXI. PROBLEM.

To cut a given right line in extreme and mean proportion.

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Let AB be the given right line; it is required to cut it in extreme and mean proportion.

Upon AB describe the fquare AC (II. 1.), and bisect the fide AD in E (I. 10.); and join BE.

In EA produced, take EF equal to EB (I. 3.); and upon AF describe the fquare FH (II. 1.); then will AB be divided at the point H as was required.

For fince DF is the fum of EB, ED, or EB, EA, and AF is their difference, the rectangle of DF, FA is equal to the difference of the fquares of EB, EA (II. 13.)

But the rectangle of DF, FA is equal to DG, because FA is equal to FG; and the difference of the fquares of EB, EA is equal to the fquare of AB (II. 14. Cor.); whence DG is equal to AC.

And if from each of thefe equals, the part AK, which is common, be taken away, the remainder AG will be equal to the remainder HC.

But equal parallelograms have the fides about equal angles reciprocally proportional (VI. 15.); whence нK is to HG as HA to HB.

And fince HK is equal to AD, or AB, and HG to HA, AB will be to HA as HA is to HB. Q. E. D.

PROP.

PROP. XXII. PROBLEM.

To divide a given right line into two such parts, that their rectangle may be equal to a given fquare, the fide of which is not greater than half the given line.

Let AB be the given line, and c the fide of the given fquare; it is required to divide AB into two fuch parts that their rectangle may be equal to the fquare of c.

Upon AB describe the femicircle BDA, and make BF perpendicular to AB (I. 11.), and equal to c (I. 3.)

Through F draw FD parallel to AB (I. 27.); and from the point D where it cuts the circle, let fall the perpendicular DE (I. 12.); and AB will be divided at £ as was required.

For fince BDA is a fimicircle (by Conft.), and DE is perpendicular to the diameter AB (by Const.), the rectangle of AE, EB will be equal to the fquare of ED (VI. 7. Car.)

But ED is equal to FB (I. 30.) or c; whence the rectangle of AE, EB will be equal to the fquare of c as was to be fhewn.

SCHOLIUM. When BF, or C, is equal to half AB, FÒ will be a tangent to the circle, and the rectangle of AE, EB will be the greateft poffible.

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