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and breadth multiplied together. And, in general, a rectangle in geometry is similar to the product of the measures of its two dimensions of length and breadth, or base and height. Also, a square is similar to, or represented by, the measure of its side multiplied by itself. So that, what is shown of such products, is to be understood of the squares and rectangles.
Corol. 3. Since the same reasoning, as in this theorem, holds for any parallelograms whatever, as well as for the rectangles, the same property belongs to all kinds of parallelograms, having equal angles, and also to triangles, which are the halves of parallelograms; namely, that if the sides about the equal angles of parallelograms, or triangles, be reciprocally proportional, the parallelograms or triangles will be equal; and, conversely, if the parallelograms or triangles be equal, their sides about the equal angles will be reciprocally proportional.
Corol. 4: Parallelograms, or triangles, having an angle in each equal, are in proportion to each other as the rectangles of the sides which are about these equal angles.
If a line be drawn in a triangle parallel to one of its sides, it will cut the other two sides proportionally.
Let DE be parallel to the side BC of the triangle ABC; then will AD: DB:: AE: EC.
For, draw BE and CD. Then the triangles DBE, DCE are equal to each other, because they have the same base DE, and are between the same parallels DE, BC (th. 25). But the two triangles ADE, BDE, on the bases AD, DB, have the same altitude; and the two triangles ADE, CDE, on the bases AE, EC, have also the same altitude; and be
cause triangles of the same altitude are to each other as their bases, therefore
the triangle ADE: BDE :: AD: DB,
But BDE is = CDE; and equals must have to equals the same ratio; therefore AD: DB:: AE: EC. Q. E. D.
Corol. Hence, also, the whole lines, AB, AC, are proportional to their corresponding proportional segments (corol. th. 66),
viz. AB: AC:: AD: AE,
A line u hich bisects any angle of a triangle, divides the opposite side into two segments, which are proportional to the two other adjacent sid s.
Let the angle ACB, of the triangle ABC, be bisected by the line CD, making the angler equal to the angles: then will the segment AD be to the segment DB, as the side AC is to the side CB. Or, AD: DB:: AC : CB.
For, let BE be parallel to CD, meeting AC produced at E. Then, because the line BC cuts the two parallels CD, BE, it makes the angle CBE equal to the alternate angle s (th. 12), and therefore also equal to the angler, which is
equal to s by the supposition. Again, because the line AE cuts the two paral lels DC, BE, it makes the angle E equal to the angle r on the same side of it (th. 14). Hence, in the triangle BCE, the angles B and E, being each equal to the angle r, are equal to each other, and consequently their opposite sides CB, CE, are also equal (th. 3).
But now, in the triangle ABE, the line CD, being drawn parallel to the side BE, cuts the two other sides AB, AE, proportionally (th. 82), making AD to DB, as is AC to CE or to its equal CB. Q. E. D.
Case 2. The proposition is also applicable when an external angle of a triangle is bisected.
Let AC, one of the sides of the triangle ABC, be produced to E, and let the angle BCE be bisected by the straight line CD, cutting AB produced in D; then
AD: DB:: AC: CB.
AD: AC::DB: CF (theor. 72);
But, BC is equal to CF; therefore,
Let BF be parallel to CD.
Then, because the line BC cuts the parallel lines CD, FB, it makes the angle CBF equal to the alternate angle BCD; and, therefore, also equal to the angle DCE, which is equal to BCD by supposition. Again, because the line EA cuts the two parallel lines CD, FB, it makes the angle DCE equal to the angle CFB, on the same side of the line. Hence, in the triangle BCF, the angle BFC, and FBC, being each equal to the angle DCE, are equal to each other; and, consequently, their opposite sides BC, CF, are also equal.
AD: DB:: AC: CB.
Now, in the triangle ADC, the line BF being drawn parallel to the side CD, cuts the two sides AD, AC, proportionally; making
For, make DG = AB, and DH = AC, and join GH. Then the two triangles ABC, DGH, having the two sides AB, AC, equal to the two DG, DH, and the contained angles A and D also equal, are identical, or equal in all respects (th. 1), namely, the angles B and C are equal to the angles G and H. But the angles B and C are equal to the angles E and F by the hypothesis; therefore also the angles G and H are equal to the angles E and F (ax. 1), and consequently the line GH is parallel to the side EF (cor. 1, th. 14).
Equiangular triangles are similar, or have their like sides proportional.
Hence then, in the triangle DEF, the line GH, being parallel to the side EF, divides the two other sides proportionally, making DG : DH :: DE : DF (cor. th. 82). But DG and DH are equal to AB and AC;. therefore also AB AC :: DE: DF. Q. E. D.
Triangles which have their sides proportional, are also equiangular.
In the two triangles ABC, DEF, if AB: DE:: AC : DF:: BC: EF; the two triangles will have their corresponding angles equal.
For, if the triangle ABC be not equiangular with the triangle DEF, suppose some other triangle, as DEG, to be equiangular with ABC. But this is impossible: for if the two triangles ABC, DEG, were equiangular, their sides would be proportional (th. 84). So that, AB being to DE as AC to DG, and AB to DE as BC to EG, it follows that DG and EG, being fourth proportionals to the same three quantities, as well as the two DF, EF, the former, DG, EG, would be equal to the latter, DF, EF. Thus, then, the two triangles, DEF, DEG, having their three sides equal, would be identical (th. 5); which is absurd, since their angles are unequal.
Triangles, which have an angle in the one equal to an angle in the other, and the sides about these angles proportional, are equiangular.
Let ABC, DEF, be two triangles, having the angle A = the angle D, and the sides AB, AC, proportional to the sides DE, DF: then will the triangle ABC be equiangular with the triangle DEF.
For, make DG = AB, and DH = AC, and join GH.
Then, the two triangles ABC, DGH, having two sides equal, and the contained angles A and D equal, are identical and equiangular (th. 1), having the angles G and H equal to the angles B and C. But, since the sides DG, DH, are proportional to the sides DE, DF, the line GH is parallel to EF (th. 82); hence the angles E and F are equal to the angles G and H (th. 14), and consequently to their equals B and C. Q. E. D. [See fig. th. 84.]
Let ABC be a right-angled triangle, and CD a perpendicular from the right angle C to the hypothenuse AB;
In a right-angled triangle, a perpendicular from the right angle, is a mean proportional between the segments of the hypothenuse; and each of the sides, about the right angle, is a mean proportional between the hypothenuse and the adjacent segment.
CD be a mean proportional between AD and DB;
Or, AD: CD:: CD: DB; and AB: BC :: BC: BDI); and AB: AC:: AC: AD.
For, the two triangles ABC, ADC, having the right angles at C and D equal, and the angle A common, have their third angles equal, and are equiangular (cor. 1, th. 17). In like manner, the two triangles ABC, BDC, having the right angles at C and D equal, and the angle B common, have their third angles equal, and are equiangular.
Hence then, all the three triangles, ABC, ADC, RDC, being equiangular, will have their like sides proportional (th. 84)
viz. AD: CD :: CD: DB;
Corol. 1. Because the angle in a semicircle is a right angle (th. 52); it follows, that if, from any point C in the periphery of the semicircle, a perpendicular be drawn to the diameter AB; and the two chords CA, CB, be drawn to the extremities of the diameter: then are AC, BC, CD, the mean proportionals as in this theorem, or (by th. 77), CD2 = AD. DB; AC2 = AB. AD; and BC2 = AB. BD.
Corol. 2. Hence AC2: BC2 :: AD: BD.
Corol. 3. Hence we have another demonstration of th. 34.
For since AC2 = AB. AD, and BC2 = AB. BD
Equiangular or similar triangles, are to each other as the squares of their like sides.
Let ABC, DEF, be two equiangular triangles, AB and DE being two like sides: then will the triangle ABC be to the triangle DEF, as the square of AB is to the square of DE, or as AB2 to DE2.
For, the triangles being similar, they have their like sides proportional (th. 84), and are to each other as the rectangles of the like pairs of their sides (cor. 4, th. 81);
Q. E. D.
therefore AB: DE:: AC: DF (th. 84),
and AB: DE:: AB: DE of equality : therefore AB: DE2 :: AB. AC: DE. DF (th. 75).
But A ABC: A DEF :: AB. AC: DE. DF (cor. 4, th. 81), therefore A ABC: A DEF:: AB2: DE
Q. E. D.
All similar figures are to each other, as the squares of their like sides.
The two figures being similar (by suppos.), they are equiangular, and have their like sides proportional (def. 67).
Let ABCDE, FGHIK, be two similar figures, inscribed in the circles whose diameters are AL and FM; then will each side AB, BC, &c. of the one figure be to the like side FG, GH, &c. of the other figure, or the whole perimeter AB + BC +&c. of the one figure, to the
Then, since the angle A is = the angle F, and the sides AB, AE, proportional to the sides FG, FK, the triangles ABE, FGK, are equiangular (th. 86). In like manner, the two triangles BCD, GHI, having the angle C = the angle H, and the sides BC, CD, proportional to the sides GH, HI, are also equiangular. Also, if from the equal angles AED, FKI, there be taken the equal angles AEB, FKG, there will remain the equals BED, GKI; and if from the equal angles CDE, HIK, be taken away the equals CDB, HIG, there will remain the equals BDE, GIK; so that the two triangles BDE, GIK, having two angles equal, are also equiângular. Hence each triangle of the one figure, is equiangular with each corresponding triangle of the other.
Therefore the A ABE: A FGK :: AB2 : FG',
But equiangular triangles are similar, and are proportional to the squares of their like sides (th. 88).
But as the two polygons are similar, their like sides are proportional, and consequently their squares also proportional; so that all the ratios AB2 to FG2, and BC2 to GH2, and DE2 to IK2, are equal among themselves, and consequently the corresponding triangles also, ABE to FGK, and BCD to GHI, and BDE to GIK, have all the same ratio, viz. that of AB2 to FG2: and hence all the antecedents, or the figure ABCDE, have to all the consequents, or the figure FGHIK, still the same ratio, viz. that of AB 2 to FG 2 (th. 72). Q. E. D.
Similar figures inscribed in circles, have their like sides, and also their whole perimeters, in the same ratio as the diameters of the circles in which they are incribed.