straight line drawn from the vertex at right angles to the equal side. Let ABC be an isosceles ▲, having the side A B E D AB AC, and let AD, drawn to AB, meet BC, produced, if it be necessary, in D; also, let BD be bisected in E: Then BC: AB:: AB: BE. For draw AE; and (S. 29. 1.) EA=EB; 8. THEOREM. The diameter of a circle is a mean proportional between the sides of an equilateral triangle and hexagon described about the circle. Let DEF be an equilateral ▲, described about the circle ABC, of which the centre is K; let the sides of the ▲ DEF touch the circle in the points A, B, C; let D, B be joined, cutting the circumference in G, and let LM be drawn touching the circle in G; so that (S. 1. 4. cor. 2.) LM is the side of a regular hexagon described about the circle ABC, and GB passes through the centre K; Then, DE: GB:: GB: LM. For join A, K; .. (E. 18. 1.) the DAK, DGL, are right, and the 4 ADK is common to the two DAK, DGL, which (S. 26. 1.) are, .'., equiangular; .•. (E. 4. 6.) DA:AK:: DG: GL: But (S. 1. 4. and cor. 1. 2.) DE is double of DA; the diameter GB is double of AK, or of DG, which (S. 1. 4. cor. 1.) is equal to AK; and LM is double of LG: ·. (E. 15. 5.) DE:GB::GB:LM. PROP. V. 9. THEOREM. Equiangular parallelograms have to one another the same ratio as the rectangles contained by the sides about equal angles in each. Let AC, DF, be two equiangular parallelograms, For draw (E. 11. 1.) BG and EI 1 to AB and DE, respectively; make BG=BC, and EI=EF; and complete the rectangles ABGH and DEIK; and produce the sides of the given, that are opposite to AB and DE, to meet AH and DK, in M and P, respectively. DEF, and And, since (hyp.) the ABC = (constr.) the ABL = 2 DEN, .. the ▲ LBC= < NEF; also (hyp.) the LCB=2NFE;... (S. 26. 1.) the two LBC, NEF, are equiangular : .. (E. 4. 1.) But (E. 1.6.) Also, BL: BC or BG::EN: EF or EI: EN: EI:: DN: DI; .. (E. 11. 1.) AL:AG:: DN: DI; = But (E. 35. 1.) AL AC; and DN = DF: ...AC: DF:: AG or ABX BC: DI or DEX EF. 10. COR. Triangles, having equal vertical angles, are to one another as the rectangles contained by the sides about those equal angles. PROP. VI. 11. THEOREM. The straight lines, drawn from the bisections of the three sides of a triangle to the opposite angles, meet in the same point. Let the sides AB, AC, of the ▲ ABC, be bi sected (E. 10. 1.) in E and F; and let BF and CE cut one another in the point R: The straight line which is drawn from A, to the bisection of BC, shall also pass through R. For join A, R, and produce AR to meet BC in D; join, also, E, F; and through R draw (E. 31. 1.) PRQ parallel to BC. And, since (constr. and E. 2. 6.) EF is parallel to BC, (E. 29. 1.) the two BFE, BRP, are equiangular; as are, also, .. (E. 2. 6.) BF: BR::CE:CR: Also (E. 4. 1.) BF:BR::EF: PR; And CE: CR:: EF: RQ; .. (E. 11. 5.) EF: PR:: EF: RQ; .. (S. 61. 1.) BD=DC; .., D is the bisection of BC; and there cannot be two straight lines joining the same two points A and D, which do not coincide; the straight line, drawn from A to the bisection of BC, passes through the point R, PROP. VII. 12. PROBLEM. To find, within a given rectilineal angle, first, the locus of all the points, from each of which, if two straight lines be drawn, to the lines containing the given angle, so as always to bè parallel to two straight lines given in position, they shall be to one another in a given ratio: And secondly, to find the locus of all the points, from each of which if two straight lines be drawn in like manner, they shall cut off from two given parts of the straight lines containing the given angle, segments that shall be to one another in a given ratio. |