THEOREM LXXXIII.

A line which bisects any angle of a triangle, divides the opposite side into two segments, which are proportional to the two other adjacent sides.

LET the angle ACB, of the triangle ABC, be bisected by the line CD, making the angle ACD equal to the angle DCB: then will the segment AD be to the segment BD, 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 DCB (th. 12), and therefore also equal to the angle ACD, which is equal to DCB by the supposition. Again, because the line AE cuts the two parallels DC, BE, it makes the angle E equal to the angle ACD on the same side of it (th. 14). Hence, in the triangle BCE, the angles B and E, being each equal to the angle ACD, 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.

THEOREM LXXXIV.

Equiangular triangles are similar, or have their like sides proportional.

LET ABC, DEF, be two equiangular triangles, having the angle A equal to the angle D, the angle B to the angle E, and consequently the angle C to the angle F; then will AB AC DE: DF.

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

B

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).

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.

THEOREM LXXXV.

Triangles which have their sides proportional, are 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

G F

B

triangles DEF, DEG, having their three sides equat, would be identical (th. 5); which is absurd, since their angles are unequal.

THEOREM LXXXVI.

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 equal to 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.

H

For, make DG equal to AB, and DH to 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.

D

E

THEOREM LXXXVII.

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.

LET ABC be a right-angled triangle, and CD a perpendicular from the right angle G to the hypothenuse AB; then will

CD be a mean proportional between AD and DB;
AC a mean proportional between AB and AD;
BC a mean proportional between AB and BD.

Or, AD: CD:: CD: DB; and AB: BC:: BC: BD; 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, BDC, being equiangular, will have their like sides proportional (th. 84); viz. AD: CD :: CD: DB; AB: AC:: AC: AD; and AB: BC :: BC: BD.

Cor. 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 (th. 77), CD2 = AD. DB; AC2 : = AB. AD; and BC2 = AB. BD.

Cor. 2. Hence AC2: BC2 :: AD: BD.

Cor. 3. Hence we have another demonstration of th. 34.
For since AC2: = AB . AD, and BC2 = AB. BD.
By addition AC2 + BC2 = AB (AD + BD) = AB2.

THEOREM LXXXVIII.

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);

theref. AB DE :: AC: DF (th. 84),

and AB : DE :: AB : DE of equality :

theref. AB2 :: DE2 :: AB. AC: DE. DF (th. 75).

But ▲ ABC: A DEF :: AB . AC: DE. DF (cor. 4, th. 81). theref. A ABC: A DEF:: AB2: DE2.

H

D

THEOREM LXXXIX.

All similar figures are to each other, as the squares of their like sides.

LET ABCDE, FGHIK, be any two similar figures, the like sides being AB, FG, and BC, GH, and so on in the same order: then will the figure ABCDE be to the figure FGHIK, as the square of AB to the square FG, or as AB2 to FG2.

H

C

A of

B

For, draw BE, BD, GK, GI, dividing the figures into an equal number of triangles, by lines from two equal angles B and G.

The two figures being similar, (hypoth.) they are equiangular, and have their like sides proportional (def. 67).

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 equiangular. Hence each triangle of the one figure is equiangular with each corresponding triangle of the other.

But equiangular triangles are similar, and are proportional to the squares of their like sides (th. 88).

Therefore A ABE: ▲ FGK:: AB2: FG2, ▲ BCD: ▲ GHI :: BC2: GH2, ▲ BDE A GIK :: DE2 : IK2.

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 FGH, 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 AB2 to FG2 (th. 72).

THEOREM XC.

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 inscribed.

LET ABCDE, FGHIK, be two similar figures, inscribed in the circles whose diameters are AL and FM; then will each side AB, BC, .... of the one figure be to the like side GF, GH, .... of the other figure, or the whole perimeter AB + BC + . of the one figure, to the whole perimeter FG+GH+....

E

CK

of the other figure, as the diameter AL to the diameter FM.

=

M

H

For, draw the two corresponding diagonals AC, FH, as also the lines BL, GM. Then, since the polygons are similar, they are equiangular, and their like sides have the same ratio (def. 67); therefore the two triangles, ABC, FGH, have the angle B = the angle G, and the sides AB, BC, proportional to the two sides FG, GH; consequently these two triangles are equiangular (th. 86), and have the angle ACB FHG. But the angle ACB = ALB, standing on the same are AB; and the angle FHG = FMG, standing on the same arc FG ; therefore the angle ALB = FMG (ax. 1). And since the angle ABL = FGM, being both right angles, because in a semicircle; therefore the two triangles ABL, FGM, having two angles equal, are equiangular; and consequently their like sides are proportional (th. 84); hence AB: FG the diameter AL: the diameter FM. In like manner, each side BC, CD, .... has to each side GH, HI,.... the same ratio of AL to FM and consequently the sums of them are still in the same ratio, viz. AB + BC + CD .... is to FG + GH + HI ........ is to diameter AL as diameter FM (th. 72).

....

THEOREM XCI.

Similar figures inscribed in circles, are to each other as the squares of the diameters

of those circles.

LET ABCDE, FGHIK, be two similar figures, inscribed in the circles whose diameters are AL and FM; then the surface of the polygon ABCDE will be to the surface of the polygon FGHIK, as AL2 to FM2. [See fig. th. 90.]

For, the figures being similar, are to each other as the squares of their like sides, AB2 to FG2 (th. 88). But, by the last theorem, the sides AB, FG, are as the diameters AL, FM; and therefore the squares of the sides AB2 to FG2, as the squares of the diameters AL2 to FM2 (th. 74). Consequently the polygons ABCDE, FGHIK, are also to each other as the squares of the diameters AL2 to FM2 (ax. 1).

THEOREM XCII.

The area of any circle is equal to the rectangle of half its circumference and half its diameter.

THE area of any circle ABD is equal to the rectangle contained by the radius, and a straight line equal to half the circumference.

If not, let the rectangle be less than the circle ABD, or equal to the circle FNH; and imagine ED drawn to touch the interior circle in F, and meet the circumference ABD in E and D. Join CD, cutting the arc of the interior circle in K. Let FH be a quadrantal arc of the inner circle, and from it take its half, from the remainder its half, and so on, until an arc FI is obtained, less than FK. Join CI, produce it to cut ED in L, and make

HT

FG = FL; so shall LG be the side of a regular polygon circumscribing the circle FNH. It is manifest that this polygon is less than the circle ABD, because it is contained within it. Because the triangle GCL is half the rectangle of base GL and altitude CF, the whole polygon of which GCL is a constituent triangle, is equal to half the rectangle whose base is the perimeter of that polygon and altitude CF. But that perimeter is less than the circumference ABD, because each portion of it, such as GL, is less than the corresponding arch of circle having radius CL, and therefore, à fortiori, less than the corresponding arch of circle with radius CA. Also CE is less than CA. Therefore the polygon of which one side is GL, is less than the rectangle whose base is half the circumference ABD and altitude CA; that is, (hyp.) less than the circle FNH, which it. contains; which is absurd. Therefore, the rectangle under the radius and half the circumference is not less than the circle ABD. And by a similar process it may be shown that it is not greater. Consequently, it is equal to that rectangle.

THEOREM XCIII.

The circumferences of circles are to each other as their radii.

B

k

THE circumferences of two circles ABD, abd, are as their radii. If possible, let the radius AC, be to the radius ac, as the circumference ABD to a circumference ihk less than abd. Draw the radius cie, and the straight line fig a chord to the circle abd, and a tangent to the circle ihk in i. From eb, a quarter of the circumference of abd, take away its half, and then the half of the remainder, and so on, until there be obtained an arc ed less than eg; and from d draw ad parallel to fg, it will be the side of a regular polygon inscribed in the circle abd, yet evidently greater than the circle ihk, because each of its constituent triangles, as acd, con