"Prop. 3. All right-angled triangles which have one of their acute angles common or equal will have equal values of n corresponding to the common or equal angles. "Let ABE, AB'E' be two right triangles, right-angled at E and E', and having the common angle A the hypothenuse of the one and leg of the other (which include the common angle A) being in the right lines A G, A F, which include the angle A, that is common to the two triangles; that is, A B and A E' are in A G, A B' and AE in AF. When the angles are equal, but not common, we may imagine ABE to be one of the triangles, and we may suppose the leg of the other triangle that is adjacent to the angle that is equal to A to be applied to AG, so that the angle which equals A shall coincide with 4; then will the hypothenuse of the applied triangle lie on AF, and we may conceive that A B' E' represents the applied triangle; so that the case of equal angles is reduced to that of a common angle. Let E C AE from E towards F, and E'C' — A E' from E' towards G, then draw right lines from C to B, and from C' to B'; and it is evident by Sim., B. I., p. 4, that A B C, ABC are isosceles triangles, BE, B'E' being the perpendiculars from their vertical angles to their bases. Join the vertices of the isosceles triangles by the right line B B'c, and put A B=BC-a, A C=b, b CE =n; also put AB' AE AB CB = B'C' a', AC = b', = n'; also let CB' = d, CB= d. From the al triangle A B B' we get, by prop. 1 (by using N, instead of n, to represent the angle A in this triangle, since n represents the angle A in the isosceles triangle ABC), c2= a2 + a2—2 Naa', (1), or if we put N=nx, we get c2: a2+ a2-2 nxa a'; and in like manner we get from the triangle B'B C, c2 = a2 + d2 + 2n x' a d, (2); where for we must use when B' is not between the points A and C, = and must be used for = when B' is between A and C, as is evident from (1) of prop. 2. Equating the two values of c2, we get, after a slight reduction, a2-d2-2n xa a' 2 nx' ad =0, (3); since 2n a= A C, and a2-d2= (a' + d) (a' — d) = (a' ±d) × AC (the upper sign being used when B' is not between A and C, and in the contrary case); hence, substituting the values of 2na and a-d2, by rejecting the common factor A C, (3) is reduced to a'd-xa' x'd 0, or a' (1-x)+d(1-x)=0, or since a'=b+d=AC±CB' (using the upper sign when B' is not between A and C, and when it is between A and C), we get AC(1-x)+CB' (1-x+1-x)=0, (4). Now it is evident that CB' must be arbitrary, and not dependent on AC or 1-x, 1-x'; .. we must have 1-x + 1 — x' = 0, and (4) is reduced to AC(1-x)=0, which, since AC is not = 0, gives 1-x=0, .. 1-x=0, or x=1, x'=1; hence (1) and (2) become c2= a2+ a2—2 na a', (1′), c2 = a2 + d2+2na d, (2′). In like manner, by regarding the angle A as belonging to the isosceles triangle A B'C', we get from the triangles A B B', B C'B', c2= a2 + a2 — 2 n'a a', (1′′), c2 = a2 + d22 + 2 n'a'd', (2′′); where for we must use when B is between A and C, and + when B is in AC produced beyond C. By equating the values of c2, as given by (1′) and (1"), we get n' =n, or as was to be proved. It is evident that A B E may represent any right triangle having A for one of its acute angles, and its hypothenuse on A G; also A B'E may denote any right triangle which has A for one of its acute angles, and its hypothenuse on AF; hence, from what has been shown, n will be the same for all the triangles represented by ABE and AB'E; that is, all right triangles which have a common or equal acute angle will have equal values of n corresponding to the common or equal acute angle. There is one case that apparently forms an exception to what has been shown; and that is when the hypothenuse of a tri ΑΕΙ AE angle that lies in one of the lines A F, A G coincides with the leg of another triangle that lies in the other of these two lines; but this exception is only apparent, for the value of n in these two triangles is the same as that of n in the two triangles A BE, ABE, .. when the hypothenuse of one triangle coincides with a leg of another triangle, the value of n, that corresponds to A in one of the triangles, is the same as in the other triangle. "Cor. We can now easily find the value of m that corresponds to the vertical angle of an isosceles triangle whose base-angles are represented by n (or to which n corresponds). = BI = BI = "For let AI be drawn from the base-angle A of the isosceles triangle ABC, perpendicular to the opposite side BC, meeting it in I, then from what has been shown we get CI= n b, or (since b 2 na) CI=2n2 a, .. BI—a— CI=a (1—2 n2), and (since by (1) of prop. 2, m = 1 — 2 n2) we get m which can also be easily obtained from other considerations. And since n corresponds equally to the base-angles of all the isosceles triangles represented by A B C, A B'C', and since m=1-2 n2, it follows that all isosceles triangles whose base-angles are equal will have equal values of m corresponding to their vertical angles. Prop. 4. If there are two right-angled triangles, such that a leg of the one divided by its hypothenuse gives the same quotient as a leg of the other divided by its hypothenuse, then shall the angle included by the leg and hypothenuse of the one triangle be equal to the angle included by the leg and hypothenuse of the other triangle. "Let ABC, DFE be two triangles right-angled at C and E, n; then shall the angle B A C equal the angle such that AC DE FDE. For on the longer leg DE of the one take DC' equal to A C, and through C' draw C'G perpendicular to D E, meeting D F in G, then shall the triangles ABC, D G C be identical; for since the right triangles DFE, D G C have the angle D common, we DC DE ᎠᎺ AC n, ... DG AB' DG = have, by prop. 3, and since DC= A C, we get D G = AB. If we take C'B' = CB, and draw a right line from D to B', the triangles A B C, D B'C' are identical, Sim., B. I., p. 4, .. D B′ A B, and of course D B': D G, which cannot be unless B' falls on G; for of the base-angles D B'G, D G B', one is acute and the other obtuse, and the same holds true whether B' is within or without the triangle DFE; hence we cannot have DB' = D G unless B' coincides with G, Sim., B. I., p. 19. Hence, since the triangles A B C, D B'C' are identical, and that B' falls on G, the triangles ABC, D G C are identical, and the angles BA C, FDE are equal, as required. m= BI AB BI B'K B'K "Cor. If we draw A K at right angles to B'C', meeting it in K (see fig., prop. 3), then since the base-angles of the isosceles triangles represented by ABC, A B'C' are equal, we have, by cor. to prop. 3, and hence the vertical angles of the isosceles triangles are equal; and the same holds true whether the perpendiculars AI, AK fall within the triangles (as in the figure) or without them; for when the perpendiculars fall without the triangles, the equality shows that the supplements of the vertical angles of the triangles are equal, and consequently the vertical angles are equal; and it is evident, by cor. to prop. 3, that the perpendiculars will both at the same time fall within or without the triangles; the case when m = 0, or BI= 0, B'K=0, is too evident to require any expla nation, for the vertical angles are evidently right. Hence the angles ABE, ABE', the halves of the vertical angles of the isosceles triangles, are equal; hence it follows that all those right-angled triangles which have an acute angle common, or equal, have their other acute angles also equal. Hence (see fig. 3) from the right triangles ABE, AB'E', having their angles B, B' equal, we get that is, if BE AB B'E' we have two (or more) right triangles which have an acute angle common or equal, then if we divide the leg of any one of them which is opposite to the (equal) angle by the leg adjacent to the angle, the quotient will equal the corresponding quotient obtained in like manner from any other (one) of the triangles; and the converse of what is here affirmed is also easily proved to be true in a manner very analogous to that given in proving the proposition above. "Prop. 5. The sum of the acute angles of a right triangle equals a right angle, and the sum of the squares of the legs equals the square of the hypothenuse. "Let ACB be the triangle, having the angle C right; from C D B draw CD, at right angles to the hypothenuse, meeting it at D; hence cor., prop. 4, since the right triangles ACD, ACB have the angle A common, their other acute angles A CD and B are equal; also since the (right) triangles B CD, ABC have a common angle, B, their other acute angles BCD and A are equal. Hence the sum of the angles A and B is equal to the sum of the angles B CD and ACD, which compose the right angle A C B, and of course the sum of the angles A and B is equal to a right angle; and consequently the sum of all the angles of the triangle ACB is equal to two right angles. Again, the right triangles ACB, ACD having the common angle A, by prop. 3, give or A C2 AB. AD; and in the same way we get from the triangles ACB, BCD, BC2 AB. BD; and consequently A C2 + B C2 = A B2, as required. AC the equality AB AD = "Cor. Since the right triangles ACD, BCD have the angles CAD, BCD equal, they (by the cor. to prop. 4) give the equality or CD2 AD. B D. CD "Prop. 6. The sum of the angles of any triangle is equal to two right angles. C "Let A CB denote any triangle; and suppose that the angle C is not less than either of the other angles of the triangle, and that the perpendicular CD is drawn from C to the opposite side AB; then, Sim., p. 17, B. I., CD will fall within the triangle ACB. Hence, since the triangles ACD, BCD B are right-angled at D, by the last prop. the sum of the acute angles A and ACD of the first of these triangles is equal to a right angle; and in the same way the sum of the acute angles B and B CD of the second triangle is equal to a right angle; but the sum of the acute angles of these triangles equals the sum of the angles of the triangle ACB ; consequently, the sum of the angles of the triangle ACB is equal to two right angles, as required. "In conclusion, we will remark that the relation of what has been |