is half a right-angle; but (constr.) the right ;.. the .*., equal to the 1 D is a DFB is half a right ▲, and is, but (constr.) BD=DE; ... DF=DE; and, in the same manner, it may be shewn that EG= DE; ... DF=DE=EG. Again, since (constr.) the D and E are right , .. (E. 28. 1.) DF is parallel to EG; and it has been shewn that DF = EG = DE; ... (E. 33. 1.) FG is equal and parallel to DE; (E. 29. 1.) the figure FDEG has all its right ; and it is equilateral; .. (E. 30. def. 1.) it is a square. PROP. XLI. 56. PROBLEM. To find a point, in either of the equal sides of a given isosceles triangle, from which, if a straight line be drawn, perpendicular to that side, so as to meet the other side produced, it shall be equal to the base of the triangle. Let ABC be the given isosceles A: It is required to find, in either of the two equal sides, as AB, a point from which if a perpendicular be drawn to AB and produced to meet AC, produced, it shall be equal to the base BC. Draw (E. 11. 1.) from B, BD 1 to AB, and make (E. 3. 1.) BD= BC; from D draw (E. 31. 1.) DE parallel to AB, meeting AC produced in E; E and from E, draw EF parallel to BD: F is the point which was to be found. For (constr.) the figure FBDE is a □; .. (E. 34. 1.) FE=BD=BC (constr.); also, since (constr.) the FBD is a right 2, the BFE is, also, (E. 29. 1.) a right . PROP. XLII. 57. THEOREM. The diameters of a parallelogram bisect each other. Let AB and CD be the diameters of the E D B ADBC; AB and CD bisect one another in the point of their intersection E. For since ADBC is a □, AD=CB (E. 34,1.) and (E. 29. 1.) the EAD of the A AED, EBC, of the ▲ BEC, and the EDA ECB; .. (E. 26. 1.) AE=EB, and DE EC. PROP. XLIII. 58. THEOREM. If in two opposite sides of a pa. rallelogram two points be assumed, one in each of those sides, equidistant from two opposite angles of the figure, and if two other points be likewise assumed, in the two other opposite sides, equidistant from the same two angles, the figure, contained by the straight lines joining the four points so assumed, shall be a parallelogram. In the opposite sides AD, BC of the ABCD, let the points E and G be taken equidistant from the opposite A and C; let also, the points F and H be taken, in the other two opposite sides, AB and DC, equidistant from A and C; and let E, F, and F, G, and G, H, and H, E, be joined: The figure EFGH is a parallelogram. For since (hyp.) AE=CG, and AFCH, and AC,... (E. 4. 1.) FE➡ that (E. 34. 1.) the 1 GH: Again since (E. 34. 1.) AB=DC, and AD=BC, and that (hyp.) of AD the part AE is equal to the part CG of BC, and of AB the part AF is equal to the part CH of DC, ... ED=GB, and DH=BF; also (E. 34. 1.) the EDH= ▲ FBG; ... (E. 4. 1.) EH = FG; and it has been proved that EF=HG; .'. (S. 18. 1.) EFGH is a parallelogram. PROP. XLIV. 59. THEOREM. If any number of parallelograms be inscribed in a given parallelogram, the diameters of all the figures shall cut one another in the same point. Let ABCD be a given, and let EFGH be anywhatever, inscribed in ABCD: The diameters of ABCD and of EFGH cut one another in the same point. For draw AC a diameter of ABCD, and FH a diameter of EFGH; let AC and FH cut one another in K; and let CB, produced, meet EF, produced, in L: Then, since AE is parallel to BC, and EF parallel to HG, the CGH (E. 29. 1.) GLE; and the GLE LEA; .. the ¿CGH = ▲ AEF; also (hyp. and E. 34. 1.) the <A=2C, and the side FE the opposite side GH, of the EFG; ... (E. 26. 1.) CHAF: Again, since the side AF of the ▲ AKF= the side CH of the ▲ CKH, and that (E. 29. 1.) the KAF, KFA are equal to the д KCH, KHC, ... (E. 26. 1.) AK= KC, and FK=KH; i. e. K is the bisection of the diameters AC, FH; .. (S. 42. 1.) all the diameters cut one another in the point K. 60. COR. From the demonstration it is manifest, that the angle contained by any two given straight lines, is equal to the angle contained by two other straight lines, that are parallel to the two given straight lines, each to each. PROP. XLV. 61. THEOREM. The diameters of an equilateral four-sided plane rectilineal figure bisect one another at right angles. Let AB and DC be the diameters of the equilateral four-sided figure ACBD, cutting one another in E: AB and DC bisect one another in E, at right angles. For, since (hyp.) ACBD is equilateral, it is (S. 18. 1.) a ; and ... (S. 42. 1.) the diameters bisect one another in E: Again, because DE= CE, and EA is common to the two AED, |