13. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides shall be either equal or supplementary, and in the former case the triangles shall be equal in all respects. Then shall the Ls ACB, DFE be either equal (as in Figs. 1 and 2) or supplementary (as in Figs. 1 and 3); and in the former case the triangles shall be equal in all respects. If the BAC the LEDF. [Figs. 1 and 2.] then the LACB=the LDFE, and the triangles are equal in all respects. I. 4. But if the BAC be not equal to the LEDF, [Figs. 1 and 3.] let the LEDF be greater than the BAC. At D in ED make the EDF' equal to the ▲ BAC. Then the A BAC, EDF' are equal in all respects. .. AC=DF'; I. 26. Hyp. I. 5. I. 13. Q.E.D. Three cases of this theorem deserve special It has been proved that if the angles ACB, DFE are not supplementary they are equal: Hence, in addition to the hypothesis of this theorem, (i) If the angles ACB, DFE opposite to the two equal sides AB, DE are both acute or both obtuse they cannot be supplementary, and are therefore equal; or if one of them is a right angle, the other must also be a right angle (whether considered as supplementary or equal to it): in either case the triangles are equal in all respects. (ii) If the two given angles are right angles or obtuse angles, it follows that the angles ACB, DFE must be both acute, and therefore equal, by (i): so that the triangles are equal in all respects. (iii) If in each triangle the side opposite the given angle is not less than the other given side; that is, if AC and DF are not less than AB and DE respectively) then the angles ACB, DFE cannot be greater than the angles ABC, DEF respectively; therefore the angles ACB, DFE are both acute; hence, as above, they are equal; and the triangles ABC, DEF are equal in all respects. II. ON INEQUALITIES. See Propositions 16, 17, 18, 19, 20, 21, 24, 25. 1. In a triangle ABC, if AC is not greater than AB, shew that any straight line drawn through the vertex A, and terminated by the base BC, is less than AB. 2. ABC is a triangle, and the vertical angle BAC is bisected by a straight line which meets the base BC in X; shew that BA is greater than BX, and CA greater than CX. Hence obtain a proof of 1. 20. 3. The perpendicular is the shortest straight line that can be drawn from a given point to a given straight line; and of others, that which is nearer to the perpendicular is less than the more remote; and two, and only two equal straight lines can be drawn from the given point to the given straight line, one on each side of the perpendicular. 4. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter. 5. The sum of the distances of any point within a triangle from its angular points is less than the perimeter of the triangle. 6. The perimeter of a quadrilateral is greater than the sum of its diagonals. 7. The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines drawn from the angular points to any given point. Prove this, and point out the exceptional case. 8. In a triangle any two sides are together greater than twice the median which bisects the remaining side. [See Def. p. 79.] [Produce the median, and complete the construction after the manner of 1. 16.] 9. In any triangle the sum of the medians is less than the peri meter. 10. In a triangle an angle is acute, obtuse, or a right angle, according as the median drawn from it is greater than, less than, or equal to half the opposite side. [See Ex. 4, p. 65.] 11. The diagonals of a rhombus are unequal. 12. If the vertical angle of a triangle is contained by unequal sides, and if from the vertex the median and the bisector of the angle are drawn, then the median lies within the angle contained by the bisector and the longer side. Let ABC be a A, in which AB is greater than AC; let AX be the median drawn from A, and AP the bisector of the vertical L BAC. Then shall AX lie between AP and AB. Produce AX to K, making XK equal to AX. Join KC. B ХР I. 4. Then the As BXA, CXK may be shewn ..the CAK is greater than the LCKA: hence AX lies within the angle BAP. I. 18. Q. E. D. 13. If the vertical angle of a triangle is contained by two unequal sides, and if from the vertex there are drawn the bisector of the vertical angle, the median, and the perpendicular to the base, the first of these lines is intermediate in position and magnitude to the other two. III. ON PARALLELS. See Propositions 27-31. 1. If a straight line meets two parallel straight lines, and the two interior angles on the same side are bisected; shew that the bisectors meet at right angles. [1. 29, 1. 32.] 2. The straight lines drawn from any point in the bisector of an angle parallel to the arms of the angle, and terminated by them, are equal; and the resulting figure is a rhombus. 3. AB and CD are two straight lines intersecting at D, and the adjacent angles so formed are bisected: if through any point X in DC a straight line_YXZ be drawn parallel to AB and meeting the bisectors in Y and Z, shew that XY is equal to XZ. 4. If two straight lines are parallel to two other straight lines, each to each; and if the acute angles contained by each pair are bisected; shew that the bisecting lines are parallel. 5. The middle point of any straight line which meets two parallel straight lines, and is terminated by them, is equidistant from the parallels. 6. A straight line drawn between two parallels and terminated by them, is bisected; shew that any other straight line passing through the middle point and terminated by the parallels, is also bisected at that point. 7. If through a point equidistant from two parallel straight lines, two straight lines are drawn cutting the parallels, the portions of the latter thus intercepted are equal. PROBLEMS. 8. AB and CD are two given straight lines, and X is a given point in AB: find a point Y in AB such that YX may be equal to the perpendicular distance of Y from CD. 9. ABC is an isosceles triangle: required to draw a straight line DE parallel to the base BC, and meeting the equal sides in D and E, so that BD, DE, EC may be all equal. 10. ABC is any triangle: required to draw a straight line DE parallel to the base BC, and meeting the other sides in D and E, so that DE may be equal to the sum of BD and CE. 11. ABC is any triangle: required to draw a straight line parallel to the base BC, and meeting the other sides in D and E, so that DE may be equal to the difference of BD and CE. See Propositions 33, 34, and the deductions from these Props. given on page 70. 1. The straight line drawn through the middle point of a side of a triangle parallel to the base, bisects the remaining side. Let ABC be a A, and Z the middle point of the side AB. Through Z, ZY is drawn par1 to BC. Then shall Y be the middle point of AC. Through Z draw ZX par1 to AC. Then in the Ao AZY, ZBX, because ZY and BC are par1, the LAZY = the L ZBX; and because ZX and AC are par1, .. also AZ ZB: I. 31. I. 29. 1. 29. Hyp. B X .. AY=ZX. I. 26. I. 34. Q. E.D. But ZXCY is a parm by construction; ZX=YC. Hence AYYC; that is, AC is bisected at Y. 2. The straight line which joins the middle points of two sides of a triangle, is parallel to the third side. Let ABC be a ▲, and Z, Y the middle points of the sides AB, AC. Then shall ZY be part to BC. Produce ZY to V, making YV equal to ZY. [A second proof of this proposition may be derived from 1. 38, 39.] |