Then the diagonals shall bisect the angles. B Proof. The two sides BA, AC are equal to the two sides DA, AC, each to each, and the base BC is equal to the base DC (hyp.); therefore the angle BAC is equal to the angle DAC (1.8). Q.E.D. 35. What is the converse of this proposition? The converse of this proposition is: Equal parallelograms upon the same base, and upon the same side of it, are between the same parallels.' Describe a parallelogram equal to a given square having an angle equal to half a right angle. Let ABCD be the given square. It is required to draw a parallelogram equal to the square ABCD, and having an angle equal to half a right angle. Draw the diagonal BD; through C draw CE parallel to BD (1. 31), and meeting AD produced in E (post. 2). Then the parallelogram DBCE shall be equal to the square ABCD, and the angle DBC shall be equal to half a right angle.. Proof. The parallelogram DBCE is equal to the parallelogram ABCD (1. 35). The diagonal BD bisects the angle ABC; therefore the angle DBC is half a right angle. Wherefore the parallelogram DBCE has been described equal to the given square ABCD, and having the angle DBC equal to half a right angle. Q.E.F. 37. Any point P is taken in the line joining an angular point A of a triangle to the middle point of the opposite side BC: prove that the triangles APB and APC are equal. Let BC be bisected in D. B D C Proof-The triangle ADB is equal to the triangle ADC (1. 38), and the triangle PDB is equal to the triangle PDC; wherefore the triangle APB is equal to the triangle APC (ax. 3). Q.E.Ď. 38. The equal sides AB, AC of an isosceles triangle are produced to D, E, so that AD = 2AB, AE=2AC. Show that CD and BE are trisected at their point of intersection. A B F E Join A with F, the point of intersection of CD, BE. Proof. In the demonstration to 1. 5, it is proved that CD is equal to BE, and that the angle FCB is equal to the angle FBC. Because the angle FBC is equal to the angle FCB, therefore FB is equal to FC (1. 6); and BA, AF are equal to CA, AF; therefore the triangles ABF, ACF are equal (1. 8). Again, because AB is equal to BD (hyp.), therefore the triangle ABF is equal to the triangle BFD (1. 38); wherefore the quadrilateral ABFC is double of the triangle BFD: but the quadrilateral ABFC is equal to the triangles ABC, FBC; therefore the triangles ABC, FBC are double of the triangle BFD: but the triangle ABC is equal to the triangle CBD (1. 38), and the triangle CBD is equal to the triangles FBD, FBC; wherefore the triangle FBD is double of the triangle FBC (ax. 3); and they are between the same parallels, therefore DF is double of FC; wherefore CF is a third of CD. Therefore CD, BE are trisected in F. Q.E.D. ABC, ABD are two on opposite sides of it. are also equal. equal triangles on the same base and If CD meets AB in E, then CE, ED B E Proof. For if CE be not equal to ED, If possible, let CE be greater than ED; and from EC cut off EF equal to ED the less (1. 3), and join FA, FB. Then, because EF is equal to ED, therefore the triangle FAE is equal to the triangle DAE, and the triangle FBE is equal to the triangle DBE (1. 38); wherefore the triangle AFB is equal to the triangle ADB (ax. 2): but the triangle ACB is equal to the triangle ADB (hyp.); therefore the triangle AFB is equal to the triangle ACB (ax. 1), the less equal to the greater, which is absurd (ax. 9). Therefore CE is not unequal to ED, that is, CE is equal to ED. Q.E.D. 39. AB, CD are parallel straight lines; AD, BC meet in E, and CD is produced to F, so that the triangles CEF, ACD are equal. Show that FB is parallel to AD. Proof-The triangle BCD is equal to the triangle ACD (1. 37), and the triangle CEF is equal to the triangle ACD (hyp.); therefore the triangle BCD is equal to the triangle CEF (ax. 1): take away from each of these equals the triangle CED; therefore the triangle BED is equal to the triangle FED (ax. 3), and these are upon the same base ED, and upon the same side of it, wherefore FB is parallel to ED, that is, AD (1. 39). Q.E.D. 40. Are the angles of these triangles equal, each to each? The angles of these triangles are equal, each to each, only when the other sides of the triangles are respectively parallel. 41. Hence deduce that two triangles formed by drawing straight lines from any point within a parallelogram to the extremities of its opposite sides are together half of the parallelogram. Let ABCD be a parallelogram, and from any point E within it let EA, EB, EC, ED be drawn. Then the triangles EAD, EBC shall be together equal to half of the parallelogram ABCD. Through E draw FG parallel to AD or BC (1. 31). Proof-The triangle AED is half of the parallelogram AFGD (1. 41), and the triangle BEC is half of the parallelogram FBCG; therefore the two triangles AED, BEC are together half of the parallelograms AFGD, FBCG, that is, half of the parallelogram ABCD. Q.E.D. PROPOSITIONS XLII.-XLVIII. (inclusive). 42. If in the figure the triangle be equilateral, and the given angle two-thirds of a right angle, the perimeters of the parallelogram and the triangle are also equal. Let the triangle ABC be equilateral, and the angle CEF two-thirds of a right angle. |