as radius, describe another circle. Draw BC a common tangent to these two circles; then ABC will be the circle required. 53. Describe with given radii two circles which shall touch each other and two straight lines which intersect. Let A B, A C be the straight lines. N B R Let M N be C M A a straight line parallel to AB at a distance equal to the first radius, and, M R a straight line parallel to A C at a distance equal to the second radius. Place a line, N R = the sum of the given radii, so that its extremities lie on MN, MR; then circles described with N and R as centres and the given radii will fulfil the required conditions. 54. With the vertices of triangles as centres, describe three circles which shall touch one another. Let a, b, c, be the sides, and let r1, r, r be the radii of the circles whose centres are respectively opposite the sides a, b, c. Then r2+r3 = a, r3+ r1 = b, r1 + r2 = c. CHAPTER XII. AREAS. SECTION I.-Equality of Areas. 88. Prove that the diagonals of a parallelogram divide it into four equivalent triangles. ▲ ADO = ▲ CDO (see fig. of Example 40), for they are on equal bases and of the same height. Similarly CDO ACB O, and ▲ ABO=AADO. 89. The diagonal A C of a parallelogram A B C D is produced to O: show that the triangles ODC, OBC are equal. Triangles ADC, CB A are equal triangles on the same base, A C, consequently Take away the equal areas ADC, CBA, and AODCA OBC. 90. Prove that if the middle point of one of the oblique sides of a trapezoid be joined to the opposite extremities of the parallel sides, the triangle formed is half the trapezoid. Let A B C D be the trapezoid, A B and DC the parallel sides, and E the middle point of B C. Join A E, D E, and A C. A DECADAC; For they are on the same base, and the altitude of the former is half that of the latter. For a similar reason ▲ ABE = AABC; hence ADEC + AABE = } A B C D. Take this sum from the whole trapezoid ; therefore ▲ AED = ABC D. 91. Two convex quadrilaterals are equal when they have one angle and their four sides equal each to each and situated in the same order. Let A B C D, E F G H be the quadrilaterals; let the sides of the first be respectively equal to those of the second; and let B and the included angle of the one respectively equal to two sides and the included angle of the other, are equal, and BD = F H. Then triangles B CD, F G H, having three sides of one respectively equal to three sides of the other, are equal. Hence the quadrilaterals are equal. 92. ABCD is a parallelogram, and E F the middle point of the sides C D and B A respectively; show that B E and D F trisect the diagonal A C. Let ABCD be a parellelogram of which the diagonal is AC; let F be the middle point of A B, and E the middle point of D C. Let DF and BE cut the diagonal AC in G and H. Through F draw F K parallel to A C. B K A G H E Since FBA B, and D E = DC; But F B and D E are also parallel ; Also since FK and AG are parallel, / BFK = FAG, therefore the triangles A FG, FB K are equal in all respects, and F K = AG. |
