PROBLEM М. To a given straight line apply a parallelogram equal to a given parallelogram. Let FG be the given straight line, and LH the given It is required to apply to FG a = LH. Produce GF, and cut off FM = HK, and on FM describe ☐ PF equal in all respects to LH. Through G draw GQ || to PM meeting PC produced in Q. Join QF and produce it to meet PM produced in S. Through S draw SVR || to MG or PQ meeting CF and QG produced in V and R. Then FR is ax, and it shall be the one required. For FR, PF are the complements about the diagonal QS of the PR, and they are ... = one another, (1.34) but PF is = LH ; ... FR is = LH and it has been described on FG. COR. Hence to a given straight line may be applied a parallelogram equal to a given triangle and having an angle equal to a given angle. For a LH may be drawn = the given and having an 2 = the given 2, and the remainder of the construction is the same as in the proposition. PROBLEM Ν. Describe a triangle equal to a given rectilineal figure. Let PQRSTV be the given rectilineal figure. It is required to describe a △ equal to it. Join PT: and draw VH || to PT, meeting ST produced in H. Join PH. Then PTH is = △ PVT; add to each the figure PQRST; (1. 32) ... figure PQRSH is = the figure PQRSTV. Similarly we may obtain a figure = RH having one fewer sides, and so on; thus we may obtain a equal to the given figure. COR. Hence a parallelogram may be drawn = a given rectilineal figure and having an angle = a given angle. SUPPLEMENT TO BOOK I. THEOREM (a). Of all straight lines which can be drawn to a given straight line from a given point without it, that which is nearer to the perpendicular is less than the more remote ; also to every straight line drawn on one side of the perpendicular there can be drawn one and only one straight line equal to it on the other side. Let AP be the perpendicular from A on BC; AQ, AR other straight lines on the same side of AP. (1. 15) Also if the figure APQRB revolve about AP, then PB will fall on PC and Q, R on points Q', R' in the line PC, and ... AQ, AR on AQ', AR' ; so that to every straight line on one side of AP an equal straight line can be drawn on the other side; and no two straight lines on the same side of the perpendicular can be equal; .. there cannot be drawn from it more than two straight lines equal to one another. THEOREM (6). If in any two sides and an opposite to one of them are respectively equal to the corresponding sides and is in another △, then shall these s be equal in certain cases. Let ABC, DEF be two as having the sides BA, AC and ABC respectively then shall △ABC be = △ DEF in certain cases. Let the ABC be applied to the DEF so that AB may be on DE and BC along EF. Then if DFE is a right z no other straight line can Hence be drawn from D=DF; .. AC must fall on DF. (I. 12) ABC is = △ DEF in all respects. line DF', and only one, can be drawn = DF; ... AC must fall on DF or DF'. Hence, if the 2 s opposite the sides AB, DE are both acute or both obtuse, then will the AS ABC, DEF be equal in all respects. DEFINITION. The locus of a point satisfying a certain geometrical condition is the line or lines composed of all possible positions of that point. PROBLEM (c). Find the locus of a point equidistant from two given points. ... Q lies on the line bisecting AB at right 4 s. Similarly every other point equidistant from A and B may be shewn to lie on the line bisecting AB at right 2 s; i.e. the locus consists of points which lie on this line. Moreover, every point in the line bisecting AB at right angles is a point of the locus. For if any point R be taken in this line, then AP, PR, and the 2 APR are respectively = BP, PR, and the 2 BPR; ... R is equidistant from A and B and is ... a point of the locus. Hence the locus of the point equidistant from A and B is the straight line bisecting AB at right 2 s. |