« PreviousContinue »
Let the angle BAC of the triangle ABC be bisected by tht straight line AD; then will K
BD : DC : : BA : AC. \N.
Through the point B draw BE parallel to DA, meeting CA produced in E. The triangle ABE is isosceles. For, since AD is parallel to EB, the angle ABE is equal to the alternate angle DAB (Prop. XXIII., B. I.); and the exterior angle CAD is equal to the interior and opposite angle AEB. But, by hypothesis, the angle DAB is equal to the angle DAC; therefore the angle ABE is equal to AEB, and the side AE to the side AB (Prop. XL, B. I.).
And because AD is drawn parallel to BE, the base of the triangle BCE (Prop. XVI.),
BD : DC : : EA : AC.
But AE is equal to AB, therefore
BD : DC : : BA : AC.
Therefore, the line, &c.
Scholium. The line which bisects the exterior angle of a triangle, divides the base produced into segments, which are proportional to the adjacent sides.
Let the line AD bisect the exterior angle CAE of the triangle ABC; then BD : DC : : BA : AC.
Through C draw CF parallel to AD; then it may be proved, as in the preceding proposition, that the angle ACF is equal to the angle AFC, and AF equal to AC. And because FC is parallel to AD (Prop. XVI., Cor. 1), BD : DC : BA : AF. But AF is equal to AC; therefore BD : DC : : BA : AC.
PROPOSITION XVIII. THEOREM.
Equiangular triangles have their homologous sides proportional, and are similar.
Let ABC, DCE be two equiangular triangles, having the angle BAC equal to the angle CDE, and the angle ABC equal to the angle DCE, and, consequently, the angle ACB equal to the angle DEC; then the homologous sides will be proportional, and we shall have
BC : CE : : BA : CD : : AC
Place the triangle DCE so that the side CE may be contiguous to BC, and in the same straight line with it; and produce the sides BA, ED till they meet in F.
Because BCE is a straight line, and the angle ACB is equal to the angle DEC, AC is parallel to EF (Prop. XXII., B. I.). Again, because the angle ABC is equal to the angle DCE, the line AB is parallel to DC; therefore the figure A CDF is a parallelogram, and, consequently, AF is equal to CD, and AC to FD (Prop. XXIX., B. I.).
And because AC is parallel to FE, one of the sides of the triangle FBE, BC : CE : : BA : AF (Prop. XVI.); but AF is equal to CD; therefore
BC : CE : : BA : CD.
Again, because CD is parallel to BF, BC : CE : : FD : DE. But FD is equal to AC; therefore
BC : CE : : AC : DE.
And, since these two proportions contain the same ratio BC : CE, we conclude (Prop. IV., B. II.) BA : CD : : AC : DE.
Therefore the equiangular triangles ABC, DCE have then homologous sides proportional; hence, by Def. 3, they are similar.
Cor. Two triangles are similar when they have two an gles equal, each to each, for then the third angles must also be equal.
Scholium. In similar triangles the homologous sides are opposite to the equal angles; thus, the angle ACB being equal to the angle DEC, the side AB is homologous to DC, and so with the other sides.
PROPOSITION XIX. THEOREM.
Two triangles which have their homologous sides proportional, are equiangular and similar.
Let the triangles ABC, DEF have their sides proportional, so that BC : EF : : AB : DE : : AC : DF; then will the triangles have their angles equal, viz.: the angle A equal to the angle D, B equal to E, and C equal to
At the point E, in the straight &
line EF, make the angle FEG equal to B, and at the point f make the angle EFG equal to C; the third angle G will be equal to the third angle A, and the \Wo triangles ABC, GEF will be equiangular (Prop. XXVII., Cor. 2, B. I.); therefore, by the preceding theorem,
BC : EF : : AB : GE. But, by hypothesis,
BC : EF : : AB : DE; therefore GE is equal to DE. Also, by the preceding theorem,
BC : EF : : AC : GF; but, by hypothesis,
BC : EF : : AC : DF; consequently, GF is equal to DF. Therefore the triangles GEF, DEF have their three sides equal, each to each; hence their angles also are equal (Prop. XV., B. I.). But, by construction, the triangle GEF is equiangular to the triangle ABC; therefore, also, the triangles DEF, ABC are equiangular and similar. Wherefore, two triangles, &c.
PROPOSITION XX. THEOREM.
Two triangles are similar, when they have an angle of the one equal to an angle of the other, and the sides containing those angles proportional.
Let the triangles ABC, DEF have the angle A of the one, equal to the angle D of the other, and let AB : DE : : AC . DF; the triangle ABC is similar to the triangle DEF.
Take AG equal to DE, also AH A. equal to DF, and join GH. Then the triangles AGH, DEF are equal, since two sides and the included angle in the one, are respectively equal to two sides and the included angle in the other (Prop. VI., B. I.). But, by hypothesis, AB : DE : : AC : DF; therefore
AB:AG::AC:AH; that is, the sides AB, AC, of the triangle ABC, are cut proportionally by the line GH; therefore GH is parallel to BC (Prop. XVI.). Hence (Prop. XXIII., B. I.) the angle AGH is equal to ABC, and the triangle AGH is similar to the triangle ABC. But the triangle DEF has been shown to be equal to the triangle AGH; hence the triangle DEF is similar to the triangle ABC. Therefore, two triangles, &o.
PROPOSITION XXI. THEOREM.
Two triangles are similar, when they have their homologous sides parallel or perpendicular to each other.
Let the triangles ABC, abc, DEF have their homologous sides parallel or perpendicular to each other ; the triangles are similar.
First. Let the homologous sides be parallel to each other. If the side AB is parallel to ab, and BC to bc, the angle B is equal to the angle b (Prop. XXVL, B. I.); also, if AC is parallel to ac, the angle C is equal to the angle c; and hence the angle A is equal to the angle a. Therefore the triangles ABC, abc are equiangular, and consequently similar.
Secondly. Let the homologous sides be perpendicular to each other. Let the side DE be perpendicular to AB, and the side DF to AC. Produce DE to I, and DF to H; then, in the quadrilateral AIDH, the two angles I and H are right angles. But the four angles of a quadrilateral are together equal to four right angles (Prop. XXVIII., Cor. 1, B. I.); therefore the two remaining angles IAH, IDH are together equal to two right angles. But the two angles EDF, IDH are together equal to two right angles (Prop. II., B. I.); therefore the angle EDF is equal to IAH or BAC.
In the same manner, if the side EF is also perpendicular to BC, it may be proved that the angle DFE is equal to C, and, consequently, the angle DEF is equal to B; hence the triangles ABC, DEF are equiangular and similar. Therefore, two triangles &c.
Scholium. When the sides of the two triangles are para.lel, the parallel sides are homologous; but when the sides are perpendicular to each other, the perpendicular sides are homologous. Thus DE is homologous to AB, DF to AC, and EF to BC
PROPOSITION XXII. THEOREM.
In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypothenuse ,
1st. The triangles on each side of the perpendicular are similar to the whole triangle and to each other.
2d. The perpendicular is a mean proportional between the segments of the hypothenuse.
3d. Each of the sides is a mean proportional between the hypothenuse and its segment adjacent to that side.
Let ABC be a right-angled triangle, having the right angle BAC, and from the angle A let AD be drawn perpendicular to the hypothenuse BC.
First. The triangles ABD, ACD are sim- B ilar to the whole triangle ABC, and to each other.
The triangles BAD, BAC have the common angle B, also the angle BAC equal to BDA, each of them being a right angle, and, therefore, the remaining angle ACB is equal to the remaining angle BAD (Prop. XXVII., Cor. 2, B. I.); therefore the triangles ABC, ABD are equiangular and similar. In like manner, it may be proved that the triangle ADC is equi angular and similar to the triangle ABC; therefore the three triangles ABC, ABD, ACD are equiangular and similar to each other.
Secondly. The perpendicular AD is a mean proportional be tween the segments BD, DC of the hypothenuse. For, sincb the triangle ABD is similar to the triangle ADC, their ho mologous sides are proportional (Def. 3), and we have BD: AD:: AD: DC.
Thirdly. Each of the sides AB, AC is a mean proportional between the hypothenuse and the segment adjacent to that side. For, since the triangle BAD is similar to the triangle BAC, we have
BC : BA : : BA : BD.
And, since the triangle ABC is similar to the triangle ACD we have
BC : CA : : CA : CD. Therefore, in a right-angled triangle, &c.
Cor. If from a point A, in the circumfe- .a
rence of a circle, two chords AB, AC are drawn to the extremities of the diameter BC, the triangle BAC will be right-angled at A (Prop. XV., Cor. 2, B. III.) ; therefore