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 = A A B' B C B' two triangles A'B'C', AB"C", are 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 proportional, and the homologous angles respectively equal. AB: A'B':: AC: A'C':: BC: B'C':, etc. &c. &c. A/ B' C Omitting the ratios which are common in these series, we have AB: A'B':: BC: B'C':: CD: C'D':: DE: D'E'::, etc. 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 B D C AB, A'B', BC, B'C' (T. VIII.). Hence, angle ACB = angle A'C'B', and 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 : P B' A' L' A B C D E 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'. THEOREM XIII. If from the right angle of a right-angled triangle a line is drawn perpendicular to the hypotenuse : I. The two partial triangles thus formed will be similar to the whole triangle, and consequently similar to each other. II. The perpendicular will be a mean proportional between the two segments of the hypotenuse. A III. Either side containing the right angle will be a mean proportional between its adjacent segment and the hypotenuse. First. The two triangles ABC, ABD are similar, since they have the common angle at B, and the angle BAC of the one equal to the angle BDA of the other, each being a right angle. For the same reason, ACB is similar to ACD, consequently these triangles are similar to each other. B D C Secondly. Comparing the two partial triangles ABD, ACD, we have BD: AD::AD: DC. Thirdly. Comparing the total triangle ABC with its partial triangle ABD, observing that in these triangles BC and AB are homologous, being the hypotenuses, that AB and BD are also homologous, being opposite the equal angles BCA, BAD, we shall obtain the proportion, BC:AB::AB:BD. The comparison of the triangles ABC, ACD will give, in like manner, BC: AC:: AC: DC. THEOREM XIV. In any right-angled triangle, the square, or second power, of the numerical value of the hypotenuse, is equal to the sum of the squares of the numerical values of the other two sides. For, taking the product of the means and extremes of the last two proportions of the preceding Theorem, we have A B D |