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But similar triangles are as the squares of their like fides (VI. 16.); whence the triangle e AB is to the triangle kry as the square of EB is to the square of KG.
And, for the same reason, the triangle EBD is to the triangle Kor as the square of EB is to the square of KG.
But ratios which are the same to the same ratio, are the same to each other (V. 11.); whence the triangle EAB is to the triangle KFG as the triangle Ebb is to the triangle kor.
And in the same manner it may be shewn that the tria angle EBD is to the triangle kui as the triangle DBC is to the triangle IGH.
'The triangle EAB, therefore, is to the triangle KFG, as the triangle EBD is to the triangle KG1, and as the triangle doc is to the triangle iGh (V. 11.)
And since the sum of the antecedents is to the sum of the consequents as the first antecedent is to its confequent (V. 16.), the polygon ABCDE will be to the polygon FGHIK as the triangle EAB is to the triangle KFG.
But the triangle BAB is to the triangle KFG as the square of AB is to the square of FG (VI. 16.); whence the polygon ABCDE is also to the polygon FGHIK as the Square of all is to the square of FG.
Q. E. D.
Parallelograms and triangles, having two equal angles, are to each other as the rectangles of the sides which are about those angles.
Let AB, ac be two parallelograms, having the angle DAF equal to the angle GAE ; then will AB be to AC as the rectangle of DA, AF is to the rectangle of GA, AE.
For let the sides 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 (1. 13.), and the angle FAE is equal to DAG (I. 15.), che angles DAF, DAG are also equal to two, right angles; whence FG is a right line (I. 14.)
And since parallelograms, of the fame altitude, are to each other as their bafes (VI. 1.), the parallelogram AB is to the parallelogram AK as Ad is to AE.
But AD is to Ae as the rectangle of AD, AF is to the rectangle of AE, AF (VI. 2. Cor. 2.); therefore AB is to AK as the rectangle of AD, AF is to the rectangle of AE, AF (V. 11.)
And in the fame manner it may be shewn, that ac is to AK as the rectangle of AC, AE is to the rectangle of
AE, AF; whence AB is to AC as the rectangle of AD, AF is to the rectangle of AG, AB (V. II and 15.)
Again, let DFA, AEG be two triangles, having the angle DAF equal to the angle GAE, then will dra be to ÅEG as the rectangle of DĀ, AF is to the rectangle of GA, AE.
For let the sides DA, AE be placed in the same 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, AE.
Q. E. D. SCHOLIUM. If the line EF be drawn, the latter part of this proposition may be proved from the triangles, independently of the former,
PRO P. XIX. THEOREM.
The rectangles under the corresponding lines of two ranks of proportionals, are thema felves proportionals.
Let as be to CB as de iš to Fe, and GH to Gk as LM to l*; then will the rectangle of Ae, oh be to that of
6B, GK as the rectangle of Des im is to that of FE, IN
For draw BQ, ES at right angles to AB; DE (I. 11.), 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 since parallelograms, of the same altitude, are to each other as their bafès (VI. 1.); AP will be to cp as AB to CB ; and DR to FR IS DE TO FE.
But AB is to CB as De to FE (by Hyp.) ; whence AP will be to 'CP as DÅ TÓ ÉR (V. 11.), óř ap to DR às CP to FR (V. 15.)
And since parallelograms of the same base are to each other as their altitudes (VI. 2.), ce will be to ce as BP to BQ; and FR to fs as Er to Ës.
Or, because BP, BQ are equal to GH, GK, and ER, ES to LM, LN (by Conjt.), cp will be to 'ce as GH to GK; and FR to is aš LM to LN (V.9.)
But oh is to GK as lM to LN (by Hyp:), therefore cp will be to coas ir is to rs (V. 11.), or CP to Fr as ce to FS (V. 15.)
And it has been before shewn thát Áp is to DR as c to Fr; whence Áp is to or as co to F's (V:11.), or AP to co as dr to Fs (V. 15.)
But Ap is the rectangle of AB, GH; ce of CB, GK; DR of DE, LM; and Fs of FE, IN (by Conft.); therefore the rectangle of AB, GH is to that of CB, GK as the redangle of DE, LM is to that of Fe, Ln.
Q. E. D. COROLL. The squares of four proportional lines are themselves proportionals,
PROP. XX. THEOREM.
The sides and diagonals of four proportional squares, are themselves proportional.
Let AC, EF, HL and on be four proportional squares; then will their fides and diagonals be also proportionals.
For, make s a fourth proportional to AB, EB and HK (VI. 9.); and draw the diagonals bo and KM.
Then, since AB is to EB as hk is to s (by Conft.), AC jwill be to ef as al is to the square of s (VI. 19. Cor.)
And because ac is to ef as he is to' On (by Hyp.), HL will be to the square of s as al is to en, or the square of
QK (V. 11.)
But magnitudes which have the fame ratio to the same magnitude are equal to each other (V. 10.); whence the square of s is equal to the square of QK.
And since equal squares have equal fides (II. 3.), s is equal to Qk; and consequently Ab is to EB as HK to to QK.
Again, because the triangles ABD, EBG are equiangular, AB will be to EB as BD TO BG (VI. 5.)
And because the triangles HKM, QKP are also equiangular, #K will be to QK, as KM to KP (VI. 5.)
But AB has been shewn to be to EB as HK is to OK ; consequently is to BG as Km is to KP (V. 11.)
Q. E. D.