the angle r, which is (by hyp.) equal to s. Again, because the line CE cuts the two parallels AD, BE, it makes the angle E equal to the angle r* on the same side (th. 10). Hence, in the triangle ABE, the angles B and E, being each equal to half the bisected angle of the triangle, are equal to each other, and, consequently, their opposite sides AB, AE are also equal (th. 4).

But now, in the triangle CBE, the line AD, being parallel to the side BE, cuts the other two sides, CB, CE, proportionally (th. 61), making CD to DB, as is CA to AE, or to its equal AB. Q. E. D.

THEOREM LXIII.

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

For, by th. 43, the rectangles of the corresponding sides taken alternately are equal, and by the second part of th. 59, these corresponding or like sidest are, in consequence, directly proportional.

THEOREM LXIV.

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.

C

G F

B

For, if the triangle ABC be not equiangular with the triangle DEF, suppose some other triangle, as DEG, constructed upon the side DE, to be equiangular with ABC. But this is impossible; for if the D two triangles ABC, DEG were equiangular, their sides would be proportional (th. 63), viz.,

E

*The use of small letters to designate angles may be adopted in other propositions.

+ The corresponding sides are called homologous.

In the same manner, it may be proved that
EG=EF:

.. (th. 5), 4* DEF is identical with A DEG, which is absurd, the angles being different.

THEOREM LXV.

Triangles which have an angle in the one equal to an angle in the other, and the sides about these angles proportional, are equiangular.

=

For, make AG = DE, and AH DF, and join GH. Then the two triangles DEF, AGH, 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 E and F. But, since the sides AG, AH are proportional to the sides. AB, AC, the line GH is parallel to BC (th. 61, corol. 2); hence the angles B and C are equal to the angles G and H respectively (th. 10), and, consequently, to their equals E and F. Q. E. ́D.

General Scholium.-Triangles will be similar, 1°. When they have their angles equal, or two of their angles equal (th. 15, corol. 1); 2°. When they have their homologous sides proportional;† 3°. When

* This sign (4) stands for the word triangle.

Triangles are the only polygons in which one part of the definition (def. 67) of similar figures involves the other as a necessary consequence. Thus a square and a rectangle are equiangular quadri

they have an angle in each equal, and the sides about the equal angles proportional; 4°. When they have their sides respectively parallel or perpendicular, or in any way equally inclined.

THEOREM LXVI.

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 AD a perpendicular from the vertex of the right angle A to the hypothenuse CB; then will

B

D C

AD be a mean proportional between BD and DC; AB a mean proportional between BC and BD; AC a mean proportional between BC and DC. For, the two right-angled triangles ABD, ABC, having the angle B common, are equiangular (cor. 2, th. 15). For a similar reason, the two triangles ABC, ADC are equiangular.

Hence, then, all the three triangles, viz., the whole triangle and the two partial triangles ABC, ABD, ADC, being equiangular, will have their like sides proportional (th. 63),

viz.,*

and

and

BD:AD::AD: DC;
BC: AC:: AC: DC;

BC: AB:: AB: BD.

Q. E. D.

laterals, but their homologous sides are not proportional, the adjacent ones of the square having a ratio of equality, those of the rectangle a ratio of inequality. Again, the sides of a square and rhombus are proportional, having in both the ratio of equality, but the angles are not equal, those of the square being right, those of the rhombus oblique.

*The student will be aided by saying BD, the long perpendicular side of the left triangle, is to AD, the long perpendicular side of the right triangle, as AD, the short perpendicular side of the former, is to DC, the short perpendicular side of the latter. Again, BC, the hypothenuse of the whole triangle, is to AB, the hypothenuse of the left partial triangle, &c.

Corol. 1. Because the angle in a semicircle is a right angle (corol. 3, th. 39), it follows that if, from any point A in the periphery of the semicircle, a perpendicular be drawn to the diameter BC, and the two chords CA, AB be drawn to the extremities of the diameter; then are AD, AB, AC the mean proportionals as in this theorem, or (by th. 55) AD' = CD. BD; AB' = BC. BD; and AC CB. CD. Corol. 2. Hence AB': AC' :: CD: BD. Corol. 3. Hence we have another demonstration of th. 26. For, since AB' BC. CD; By addition, AB2 + AC2 = BC (BD + CD) = BC'.

=

BC. BD, and AC'

=

THEOREM LXVII.

Similar triangles are to each other as the squares of their like sides.

Let ABC, DEF be two A similar triangles, AB and DE being two like sides; then will the triangle ABC be to the triangle DEF as G the square of AB is to the square of DE, or as AB2 to DE'.

B

D

H

C E

For, the triangles being similar, they have their like sides proportional (def. 67);

therefore

AB: DE:: AC: DF;

and AB: DE:: AB: DE, an identity of ratios; therefore AB': DE':: AB. AC: DE. DF (th. 53). But the triangles are to each other as the rectangles of the like pairs of their sides (cor. 3, th. 60); or A ABC: A DEF:: AB. AC: DE.DF;

therefore

▲ ABC: A DEF:: AB2: DE'. Q. E. D.

THEOREM LXVIII.

The perimeters of all similar figures are to each other as their homologous sides, and the surfaces as the squares of their homologous sides.

will the perimeter of the figure ABCDE be to the perimeter of the figure FGHIK as AB to FG, and the surface as the square of AB to the square of FG, or as AB' to FG2.

For (by def. 67) AB: BC: CD, &c. :: FG: GH: HI, &c. And (by th. 50) AB + BC + CD, &c.; or the perimeter of the first polygon is to FG+ GH + HI, &c.; or the perimeter of the second polygon as AB: FG.

Again, draw AC, AD, FH, FI, dividing the figures into an equal number of triangles by lines from two equal angles A and F.

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

=

Then, since the angle B is the angle G, and the sides AB, BC proportional to the sides FG, GH, the triangles ABC, FGH are equiangular (th. 65). If, from the equal angles BCD, GHI there be taken the equal angles ACB, GHF, there will remain the equals ACD, FHI; and since, from the similarity of the triangles ABC, FGH, and of the whole polygons, AC and FH, as well as CD and HI, have the same ratio that BC and GH have, they must have the same ratio as one another; hence the triangles ACD, FHI, having an equal angle contained by proportional sides are (th. 65) similar.

In the same manner, ADE may be proved similar to FIK. Hence each triangle of the one figure is equiangular with each corresponding triangle of the other. But equiangular triangles are similar (th. 63), and are proportional to the squares of their like sides (th. 67).