PROPOSITION XI. PROBLEM. To draw a straight line at right angles to a given straight line from a given point in the same. Let AB be the given st. line, and C a given pt. in it. Take any pt. D in AC, and in CB make CE=CD. Join FC. FC shall be 1 to AB. I. 1. For in As DCF, ECF, :: DC=CE, and CF is common, and FD=FE, .. 4 DCFL ECF; and .. FC is 1 to AB. I. c. Def. 9. Q. E. F. COR. To draw a straight line at right angles to a given straight line AC from one extremity, C, take any point D in AC, produce AC to E, making CE=CD, and proceed as in the proposition. Ex. 1. Shew that in the diagram of Prop. Ix. AF and ED intersect each other at right angles, and that ED is bisected by AF. Ex. 2. If O be the point in which two lines, bisecting AB and AC, two sides of an equilateral triangle, at right angles, meet; shew that OA, OB, OC are all equal. Ex. 3. Shew that Prop. XI. is a particular case of Prop. IX. PROPOSITION XII. PROBLEM. To draw a straight line perpendicular to a given straight line of an unlimited length from a given point without it. A D Let AB be the given st. line of unlimited length; the given pt. without it. “ It is required to draw from C a st. line 1 to AB. Take any pt. D on the other side of AB. With centre C and distance CD describe a cutting AB in E and F. Bisect EF in O, and join CE, CO, CF. Then CO shall be to AB. I. 10. For in AS COE, COF, · EO=FO, and CO is common, and CE=CF, .. CO is to AB. I. c. Def. 9. Q. E. F. Ex. 1. If the straight line were not of unlimited length, how might the construction fail? Ex. 2. If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles. Ex. 3. The lines drawn from the angular points of an equilateral triangle to the middle points of the opposite sides are equal. Miscellaneous Exercises on Props. I. to XII. 1. Draw a figure for Prop. II. for the case when the given point A is (a) below the line BC and to the right of it. (B) below the line BC and to the left of it. 2. Divide a given angle into four equal parts. 3. The angles B, C, at the base of an isosceles triangle, are bisected by the straight lines BD, CD, meeting in D; shew that BDC is an isosceles triangle. 4. D, E, F are points taken in the sides BC, CA, AB, of an equilateral triangle, so that BD=CE=AF. Shew that the triangle DEF is equilateral. 5. In a given straight line find a point equidistant from two given points; 1st, on the same side of it; 2d, on opposite sides of it. 6. ABC is a triangle having the angle ABC acute. In BA, or BA produced, find a point D such that BD=CD. 7. The equal sides AB, AC, of an isosceles triangle ABC are produced to points F and G, so that AF=AG. BG and CF are joined, and H is the point of their intersection. Prove that BH=CH, and also that the angle at A is bisected by AH. 8. BAC, BDC are isosceles triangles, standing on opposite sides of the same base BC. Prove that the straight line from A to D bisects BC at right angles. 9. In how many directions may the line AE be drawn in Prop. III. ? 10. The two sides of a triangle being produced, if the angles on the other side of the base be equal, shew that the triangle is isosceles. 11. ABC, ABD are two triangles on the same base AB and on the same side of it, the vertex of each triangle being outside the other. If AC=AD, shew that BC cannot =BD. 12. From Cany point in a straight line AB, CD is drawn at right angles to AB, meeting a circle described with centre A and distance AB in D; and from AD, AE is cut off = AC: shew that AEB is a right angle. PROPOSITION XIII. THEOREM. The angles which one straight line makes with another upon one side of it are either two right angles, or together equal to two right angles. Let AB make with CD upon one side of it the 4s ABC, ABD. Then must these be either two rt. Ls, or together equal to two rt. 4 s. First, if ▲ ABC= < ABD as in Fig. 1, each of them is a rt. 4. Secondly, if ABC be not= ▲ ABD, as in Fig. 2, from B draw BE L to CD. Def. 9. I. 11. Then sum of 4s ABC, ABD=sum of 4 s EBC, EBA, ABD, and sum of s EBC, EBD=sum of 4 s EBC, EBA, ABD; .. sum of 4 s ABC, ABD=sum of 4 s EBC, EBD ; Ax. 1. .. sum of 4 s ABC, ABD=sum of a rt. ▲ and a rt. ≤ ; .. 4s ABC, ABD are together two rt. 4 s. Q. E. D. Ex. Straight lines drawn connecting the opposite angular points of a quadrilateral figure intersect each other in O. Shew that the angles at O are together equal to four right angles. NOTE (1.) If two angles together make up a right angle, each is called the COMPLEMENT of the other. Thus, in fig. 2, 4 ABD is the complement of ▲ ABE. (2.) If two angles together make up two right angles, each is called the SUPPLEMENT of the other. Thus, in both figures, ▲ ABD is the supplement of ABC. PROPOSITION XIV. THEOREM. If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines must be in one and the same straight line. At the pt. B in the st. line AB let the st. lines BC, BD, on opposite sides of AB, make 2 s ABC, ABD together=two rt. angles. Then BD must be in the same st. line with BC. For if not, let BE be in the same st. line with BC. Ls ABC, ABE together=two rt. 4 s. Then And 4s ABC, ABD together=two rt. 4 s. .. sum of 4s ABC, ABE=sum of 4 s ABC, ABD. I. 13. Hyp. Take away from each of these equals the ABC; Ax. 3. then 4 ABEL ABD, that is, the less-the greater; which is impossible, .. BE is not in the same st. line with BC. Similarly it may be shewn that no other line but BD is in the same st. line with BC. .. BD is in the same st. line with BC. Q. E. D. Ex. Shew the necessity of the words the opposite sides in the enunciation. |