PROP. 12. PROB. To draw a straight line at right angles to a given straight line, from a given point without it. Given st. line AB, and point C without it. To draw from C a st. line AB. With centre C, and any two points E and D (Post. 3). Bisect ED at H (I. 10). Join CH (Post. 1). CH is the line required. Join CE and CD. In triangles CHE, CHD, EH = HD (cons.), and CH is common, and CE = CD (Def. 11), .. ^EHC= ^DHC (I. 8), and .. EHC- = ▲ DHC, .. CH is AB (Def. 7), and is drawn from the given point C. Q.E.F. 23. If the given line could not be produced, show how the above construction might fail. 24. From a given point without a given straight line two equal lines can be drawn to the given straight line, one on each side of the perpendicular. 25. Given two points on opposite sides of a straight line, and at unequal distances from it, to draw from these points two straight lines to meet in the given straight line, and include an angle bisected by the given straight line. 26. If the middle point of a chord of a circle be joined with the centre, this line is perpendicular to the chord. A chord of a circle is a straight line joining any two points on the circumference. PROP. 13. THEOR. The angles which one straight line makes with another on one side of it are either two right angles, or are together equal to two right angles. Given AB meeting CD and making with its. ABC and ABD. To prove ABC and ABD either two right angles, or together equal to two right angles. ABD (as in Fig. 1), each is a right angle (Def. 7). But if ABC be not = ABD (as in Fig. 2), from B But angles DBE, EBC, are right angles (cons.); ▲ CBA and ABD together = 2 Ls. Q.E.D. Note. If two angles together equal two right angles, each is called the supplement of the other. ABD and ABC are supplementary angles. If two angles together equal one right angle, each is called the complement of the other. In Fig. 2, ABD and ABE are complementary angles. 27. If two lines intersect, the four angles at the point of intersection are together equal to four right angles. 28. If any number of lines meet at a point, the angles made by these lines taken consecutively, amount to four right angles. PROP. 14. THEOR. If, at a point in a straight line, two other straight lines, on the opposite sides of it, make the adjacent angles together equal to two right angles, the two straight lines shall be in one and the same straight line. If BD be not in the same st. line with CB, produce CB to E (Post. 2), and BE will not coincide with BD. AB meets the st. line CBE, ≤ ABC + ▲ ABE 2 Ls. (I. 13). But ABC + ABD 2 Ls. (hyp.); = .. ABC + 2 ABE = < ABC + ▲ ABD (Ax. 1 and 10). Take ABC from each, :. ≤ ABE = 2 ABD (Ax. 3), the less the greater, which is impossible. .. BE and BD must coincide, that is, BD is in the same st. line with CB. Q.E.D. 29. If four straight lines meet at a point and make the opposite angles equal to one another, the lines are two and two in the same straight line. 30. If two supplementary angles are bisected, the bisecting lines are at right angles to one another. 31. The bisectors of two conjugate angles are in the same straight line. Note. When two straight lines meet at a point, they form two angles, which together amount to 4 s.; these angles are said to be conjugate. The smaller of the two conjugate angles is to be understood as the one meant when the reference is simply to the angle made by two lines. 32. To prove that there is only one perpendicular to a straight line from a point outside of it. PROP. 15. THEOR. If two straight lines cut one another, the opposite vertical angles shall be equal. Given AB and CD inter secting at the point E. To prove AEC = = L BED, also AED = 4 BEC. AE meets st. line CD, C LAEC+ AED (I. 13). = 2 L's. Again, DE meets st. line AB, ▲ AED + ▲ DEB = 2 L''s. (I. 13). . AEC + 2 AED = AED + DEB (Ax. 1). Take away the common angle AED, .4 AEC: ▲ DEB (Ax. 3). = Similarly, AED = CEB. Q.E.D. Another proof of this proposition may be given as follows: Let the line AE be made to revolve round the point E into the position of CE, then EB will at the same time revolve into position of DE. .. whatever amount of turning is required to bring AE into position of CE, a like amount of turning will bring BE into position of DE; .. 4 AEC 4 BED. Similarly, AED=✩ BEC. Q.E.D. 33. If the lines AB and CD in I. 15 bisect one another, the triangles formed by joining AC and BD are equal in every respect. 34. If the diagonals of a quadrilateral bisect one another, the quadrilateral has its opposite sides equal. 35. The bisectors of two opposite vertical angles are in the same straight line. PROP. 16. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. .. AABE = B = E ▲ FEC (I. 15), ▲CFE (I. 4), and A = ECF. But ECD is > ECF, . ECD, that is, ACD, ▲ ▲ ▲ is > A. In a like way, by producing AC to G, it can be proved that BCG is > 2 ABC; but BCG = 4 ACD ABC. Q.E.D. 36. In the Fig. to I. 16, in what respect does the triangle FBC equal the triangle ABC ? 37. If the angles at the base of a triangle are both acute, the perpendicular from the vertex on the base must fall within the triangle. 38. If one of the angles at the base of a triangle is an obtuse angle, the perpendicular from the vertex on the base must fall without the triangle, towards the side on which is the obtuse angle. |