The homologous sides, AB and A'B', AC and A'C', BC and B'C', as we see, are opposite the equal angles C'= C, B'= B, A'=A.. THEOREM VII. Two triangles are similar when the sides of the one are respectively parallel or perpendicular to the sides of the other, and the homologous sides are the parallel or the perpendicular sides. From the preceding Theorem we see that the demonstration will be made out if we can show that these triangles have their angles respectively equal. Let AB and A'B', AC and A'C', BC and B'C', represent the respective sides, either parallel or perpendicular, we are to show that C=C', B=B', A=A'. In effect, we know (B. I., T. III., C.) that the angles C and C', B and B', A and A', must be either equal or supplementary. Now, in the first place, it is impossible that three set of angles should be supplementary, since we should then have A+ A+B+B' + C+C'= 6 right angles, which is absurd (B. I., T. V., C. I.). Neither can two set of these angles be supplementary; as, for example, A and A', B and B', for then we should have A+A'+B+B'= 4 right angles, which is also absurd. It is then necessary that two angles, at least, of the first triangle should be equal respectively to two angles of the second; consequently, the third angle of the first triangle must equal the third angle of the second (B. I., T. V., C. III.). So that we shall have A = A', B = B', C = C'; and consequently the triangles are similar, and we have this series of equal ratios (T. VI.): BC: B'C':: AC: A'C':: AB: A'B'. From the method of demonstration we see that the homologous sides BC and B'C', AC and A'C', AB and A'B', are the parallel or the perpendicular sides. THEOREM VIII. Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing these angles proportional, are similar. In the two triangles ABC, A'B'C', suppose we have A = A', and AB: A'B':: AC: A'C'. Take on AB and AC, AB" = A'B', AC" A'C', and draw B"C". The = two triangles A'B'C', AB"C", are A B' C" equal, having an equal angle included between equal sides each to each. Now, by hypothesis, we have consequently, AB: A'B':: AC: A'C', AB: AB":: AC: AC"; and B"C" is parallel to BC (T. III., Cor. II.), and the triangle AB"C" is therefore similar to ABC (T. IV.). Then, also, is A'B'C' similar to ABC. SIMILAR POLYGONS. THEOREM IX. Two similar polygons have their homologous sides proportion al, and the homologous angles respectively equal. AB: A'B':: AC: A'C':: BC: B'C'::, etc. &c. &c. Omitting the ratios which are common in these series, we have AB: A'B':: BC: B'C':: CD: C'D':: DE: D'E': Again, from the similarity of these same triangles, it follows that their homologous angles are equal (T. V.), and consequently the respective angles of the two polygons are equal, since they are composed of angles equal each to each. Thus, A = A', B=B', C=C', D=D', etc. THEOREM X. Two polygons are similar when they have their sides, taken in the same order, proportional, and their angles, also taken in the same order, equal each to each. That is to say, when we have and AB: A'B':: BC: B'C':: CD: C'D'::, etc., A=A', B=B', C=C', D=D', etc. Let the two polygons be decomposed into triangles by drawing lines from A and A'. The two triangles ABC, F A'B'C', are similar, having E F E' = angle AB, A'B', BC, B'C' (T. VIII.). Hence, angle ACB = AC: A'C'::BC: B'C'. But, by hypothesis, we have BC: B'C':: CD: C'D'. Consequently, AC: A'C':: CD: C'D'. Now, comparing the two triangles ACD, A'C'D', we see that the angles ACD, A'C'D' are equal, being the differences between the equal angles BCD, B'C'D', and ACB, A'C'B', and we have just shown that the sides about these equal angles are proportional, hence these triangles are also similar. The same may be shown for all the couples of triangles, ADE and A'D'E', AEF and A'E'F'. Hence the polygons are similar (D. III.). Scholium. Two parallelograms are similar, when they have an angle of the one equal to an angle of the other, and the sides containing the equal angles proportional. Two rhombuses are similar, when they have an angle of the one equal to an angle of the other. All squares are similar figures. All regular polygons of the same number of sides are similar figures. PROPORTIONAL LINES.-PROPERTIES OF THE SIDES OF TRIANGLES. THEOREM XI. The intercepted portions of two parallel lines, made by any number of lines proceeding from the same point, are proportional. The comparison of the different couples of similar triangles PAB and PA'B', PBC and PB'C', etc., immediately gives the following series of equal ratios: A' A B P B' CADA EL L' C D E L Now, all the ratios are equal, however far they may be extended, since the last in each line above is the first of the following line. Then, if we use only the intermediate ratios, we shall have AB: A'B':: BC: B'C' :: CD: C'D':: DE: D'E' ::, etc., which establishes the proposition. When the point P is within the parallels, the same demonstration will apply. In this case, the corresponding segments of the two lines AL and A'L' will be situated in opposite directions. THEOREM XII. If a line be drawn bisecting either angle of a triangle, it will divide the opposite side into two segments proportional to their adjacent sides. since they are equiangular; hence AB:AC::BE: CF. But the two triangles BDE, CFD are also similar, and give Consequently, by equality of ratios, we have AB: AC:: BD: CD. Cor. The line AD', bisecting the supplementary angle at A, that is to say, the line perpendicular to AD, will also determine, on the prolongation of the side BC, two segments BD', CD', such as to give the proportion, AB: AC:: BD': CD'. For, drawing BE' and CF' perpendicular to AD', we have, by reason of similar triangles, AB: AC:: BE': CF'; BE': CF' : : BD': CD'; consequently, AB: AC:: BD': CD'. Scholium. From the two proportions, AB: AC:: BD : CD, AB: AC:: BD' : CD', just demonstrated, we deduce the following: BD: CD:: BD': CD'; and the straight line BC is said to be divided harmonically at the points D and D'. |