But AB is common to the two triangles; therefore, the three sides of the ▲ ABD are respectively equal to the three sides of the ▲ ABC, and the two A's are equal, (B. I, Th. 21). But the A's ABD, and abc, are equiangular by construction; therefore, the A's, ABC, and abc, are also equiangular and similar. Hence the theorem; if any triangle have its sides, etc. Second Demonstration. Let abc and ABC be two triangles whose sides are respectively proportional, then will the triangles be equiangular and similar. That is, LaLA,L6=LB, and Lc=LC. If the c be in fact equal to the LC, the triangle abc can be placed on the triangle ABC, ca taking the direction of CA and cb of CB. The α line ab will then divide A B the sides CA and CB proportionally, and will therefore be parallel to AB, and the triangles will be equiangular and similar, (Th. 17). But if the c be not equal to the LC, then place ac on AC as before, the point c falling on C. Under the present supposition cb will not fall on CB, but will take another direction, CV, on one side or the other of CB. Make CV equal to cb and draw a V. Now, the ▲ abc is represented in magnitude and position by the ▲ a VC; and if, through the point a, the line ab be drawn parallel to AB, we shall have CA: ab : AB; Ca but by (Hy.) Hence, (Th. 6), ab AB aV: AB; which requires that ab = = a V, but (Th. 22, B. I) ab can not be equal to a V; hence the last proportion is absurd, and the supposition that the Lc is not equal to the LC, which leads to this result, is also absurd. Therefore, thee is equal to the LC, and the triangles are equiangular and similar. Hence the theorem; if any triangle have its sides, etc. THEOREM XIX. If four straight lines are in proportion, the rectangle contained by the lines which constitute the extremes, is equivalent to that contained by those which constitute the means of the proportion. Let A, B, C, D, represent the four lines; then we are to show, geometrically, that A x D B x C. = Al B D Place A and B at right angles to each other, and draw the hypotenuse. Also place C and D at right angles to each other, and draw the hypotenuse. Then bring the two triangles together, so that C shall be at right angles to B, as represented in the figure. F B BC C D AD Now, these two A's have each a R. L, and the sides about the equal angles are proportional; that is, A: B:: C: D; hence, (Th. 17, Cor. 2), the two A's are equiangular, and the acute angles which meet at the extremities of B and C, are together equal to one right angle, and the lines B and C are so placed as to make another right angle; therefore, also, the extremities of A, B, C, and D, are in one right line, (Th. 3, B. I), and that line is the diag onal of the parallelogram bc. By Th. 31, B. I, the complementary parallelograms about this diagonal are equal; but, one of these parallelograms is B in length, and C in width, and the other is D in length and A in width; therefore, Hence the theorem; if four straight lines are in proportion, etc. = = Cor. When B C, then A x D B2, and B is the mean proportional between A and D. That is, if three straight lines are in proportion, the rectangle contained by the first and third lines is equivalent to the square described on the second line. THEOREM XX. Similar triangles are to one another as the squares of their homologous sides. Let ABC and DEF be two similar triangles, and LC and MF perpendiculars to the sides. AB and DE respectively. Then we are to prove that AABC: A DEF AB': DE'. = : АВ By the similarity of the triangles, we have, B LC: MF = AB: DE = A D H I E AB x LC: DE × MF. But, (by Th. 30, B. I), AB × LC is double the area of the ▲ ABC, and DE × MF is double the area of the A DEF. Therefore, AABC: A DEF:: AB × LC : DEXMF And, (Th. 6), ▲ ABC: A DEF= AB2: DE2. etc. Hence the theorem; similar triangles are to one another, The following illustration will enable the learner fully to comprehend this important theorem, and it will also serve to impress it upon his memory. Let abc and ABC represent two equiangular triangles. Suppose the length of the side ac to be two units, and the length of the corresponding side AC to be three units. Now, drawing lines through the points of a b A C B division of the sides ac and AC, parallel to the other sides of the triangles, we see that the smaller triangle is composed of four equal triangles, while the larger contains nine such triangles. That is, the sides of the triangles are as 2: 3, and their areas are as 4:92' 3'. THEOREM XXI. Similar polygons may be divided into the same number of triangles; and to each triangle in one of the polygons there will be a corresponding triangle in the other polygon, these triangles being similar and similarly situated. Let ABCDE and abcde be two similar polygons. Now it is obvious that we can divide each polygon into as many triangles as the figure has sides, less E B two; and as the polygons have the same number of sides, the diagonals drawn from the vertices of the homologous angles will divide them into the same number of triangles. C Since the polygons are similar, the angles EAB and eab, are equal, and EA : AB :: ea: ab. Hence the two triangles, EAB and eab, having an angle in the one equal to an angle in the other, and the sides about these angles proportional, are equiangular and similar, and the angles ABE and abe are equal. But the angles ABC and abc are equal, because the polygons are similar. Hence, LABC—LABE=| abc—Labe; that is, EBC = | ebc. The triangles, EAB and eab, being similar, their homologous sides give the proportion, and since the polygons are similar, the sides about the equal angles B and b are proportional, and we have or, AB: BC: ab: be; BC: AB:: bc: ab. (2) Multiplying proportions (1) and (2), term by term, and omitting in the result the factor AB common to the terms of the first couplet, and the factor ab common to the terms of the second, we have BC: BE:: bc: be. Hence the A's EBC and ebc are equiangular and similar; and thus we may compare all of the triangles of one polygon with those like placed in the other. Hence the theorem; similar polygons may be divided, etc. THEOREM XXII. The perimeters of similar polygons are to one another as their homologous sides; and their areas are to one another as the squares of their homologous sides. Let ABCDE and abcde be two similar polygons; then we are to prove that AB is to the sum of all the sides |