PROPOSITION 11. 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 straight line, and a the given point in it. It is required to draw from C a straight line at right angles to AB. Construction. In AC take any point D, and from CB cut off CE equal to CD. I. 3. On De describe the equilateral triangle DFE. I. 1. Join CF. Constr. and CF is common to both; Because and the third side DF is equal to the third side EF: Def. 24. therefore the angle DCF is equal to the angle ECF: 1. 8. and these are adjacent angles. But when one straight line, standing on another, makes the adjacent angles equal, each of these angles is called a right angle; Def. 10. therefore each of the angles DCF, ECF is a right angle. Therefore CF is at right angles to AB, and has been drawn from a point C in it. Q.E.F. EXERCISE. In the figure of the above proposition, shew_that any point in FC, or FC produced, is equidistant from D and E. PROPOSITION 12. PROBLEM. To draw a straight line perpendicular to a giren straight line of unlimited length, from a given point without it. Let AB be the given straight line of unlimited length, and let C be the given point without it. It is required to draw from C a straight line perpendicular to AB. Construction. On the side of AB remote from C take any point D; and with centre C, and radius CD, describe the circle FDG, cutting AB at F and G Bisect FG at H; 1. 10. Join CF and CG. Then in the triangles FHC, GHC, Constr. and HC is common to both; Because and the third side CF is equal to the third side CG, being radii of the circle FDG; Def. 15. therefore the angle CHF is equal to the angle CHG; 1. 8. and these are adjacent angles. But when one straight line, standing on another, makes the adjacent angles equal, each of these angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. Def. 10. Therefore CH is perpendicular to AB, and has been drawn from the point C without it. Q.E.F. NOTE. The line AB must be of unlimited length, that is, capable of production to an indefinite length in either direction, to ensure its being intersected in two points by the circle FDG. QUESTIONS AND EXERCISES FOR REVISION. 1. Distinguish between a problem and a theorem. 2. When are two figures said to be identically equal ? Under what conditions has it so far been proved that two triangles are identically equal ? 3. Explain the method of proof known as Reductio ad Absurdum. Quote the enunciations of the propositions in which this method has so far been used. 4. Quote the corollaries of Propositions 5 and 6, and shew that each is the converse of the other. 5. What is meant by saying that Euclid's reasoning is deductive ? Shew, for instance, that the proof of Proposition 5 is a deductive argument. 6. Two forts defend the mouth of a river, one on each side ; the forts are 4000 yards apart, and their guns have a range of 3000 yards. Taking one inch to represent a length of 1000 yards, draw a diagram shewing what part of the river is exposed to the fire of both forts. 7. Define the perimeter of a rectilineal figure. A square and an equilateral triangle each have a perimeter of 3 feet: compare the lengths of their sides. a 8. Shew how to draw a rhombus each of whose sides is equal to a given straight line PQ, which is also to be one diagonal of the figure. 9. A and B are two given points. Shew how to draw a rhomvus having A and B as opposite vertices, and having each side equal to a given line PQ. Is this always possible ? 10. Two circles are described with the same centre 0; and two radii OA, OB are drawn to the inner circle, and produced to cut the outer circle at D and E : prove that (i) DB=EA; EXERCISES ON PROPOSITIONS 1 TO 12. 1. Shew that the straight line which joins the vertex of an isosceles triangle to the middle point of the base is perpendicular to the base. 2. Shew that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides, are equal to one another. 3. Two given points in the base of an isosceles triangle are equi. distant from the extremities of the base : shew that they are also equidistant from the vertex. 4. If the opposite sides of a quadrilateral are equal, shew that the opposite angles are also equal. 5. Any two isosceles triangles XAB, YAB stand on the same base AB : shew that the angle XAY is equal to the angle XBY ; and if XY be joined, that the angle AXY is equal to the angle BXY. 6. Shew that the opposite angles of a rhombus are bisected by the diagonal which joins them. 7. Shew that the straight lines which bisect the base angles of an isosceles triangle form with the base a triangle which is also isosceles. 8. ABC is an isosceles triangle having AB equal to AC; and the angles at B and C are bisected by straight lines which meet at O: shew that OA bisccts the angle BÃC. 9. Shew that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral. 10. The equal sides BA, CA of an isosceles triangle BAC are produced beyond the vertex Á to the points E and F, so that AE is equal to ÅF ; and FB, EC are joined: shew that FB is equal to EC. 11. Shew that the diagonals of a rhombus bisect one another at right angles. 12. In the equal sides AB, AC of an isosceles triangle ABC two points X and Y are taken, so that AX is equal to AY ; and CX and BY are drawn intersecting in 0 : shew that (i) the triangle BOC is isosceles ; (iii) AO, if produced, bisects BC at right angles. 13. Describe an isosceles triangle, having given the base and the length of the perpendicular drawn from the vertex to the base. 14. In a given straight line find a point that is equidistant from two given points. In what case is this impossible ? H.S.E. PROPOSITION 13. THEOREM. The adjacent angles which one straight line makes with another straight line, on one side of it, are either two right angles, or are together equal to two right angles. Proof. Let the straight line AB meet the straight line DC. Then the adjacent angles DBA, ABC shall be either two right angles, or together equal to two right angles. CASE I. For if the angle DBA is equal to the angle ABC, each of them is a right angle. Def. 10. CASE II. But if the angle DBA is not equal to the angle ABC, from B draw BE at right angles to CD. 1. 11. Now the angle DBA is made up of the two angles DBE, EBA; to each of these equals add the angle ABC; then the two angles DBA, ABC are together equal to the three angles DBE, EBA, ABC. Ax. 2. Again, the angle EBC is made up of the two angles EBA, ABC; to each of these equals add the angle DBE; then the two angles DBE, EBC are together equal to the three angles DBE, EBA, ABC. Ax. 2. But the two angles DBA, ABC have been shewn to be equal to the same three angles; therefore the angles DBA, ABC are together equal to the angles DBE, EBC. Ax. 1. But the angles DBE, EBC are two right angles ; Constr. therefore the angles DBA, ABC are together equal to two right angles. Q.E.D. |