DC is equal to the squares on DA and AC (I. 47), because DAC is a right angle; and the square on BC is given equal to the squares on BA and AC; therefore the square on DC is equal to the square on BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC and BAC, and the base DC likewise equal to the base BC, the angle DAC is equal to the angle BAC (I. 8). But DAC is a right angle; therefore also BAC is a right angle. COR.-If the sides of a triangle are 3, 4, and 5, taken from any scale of equal parts, it is a right-angled triangle. For 52 = 32 + 42, or 25 = 9 + 16. PROPOSITION C. THEOREM. If two triangles have two sides and an angle opposite to one of them in the one, equal respectively to two sides and the corresponding angle of the other, and if the other two angles opposite to the other two equal sides are of the same species, then the triangles are equal in every respect. First, Given the triangles ACD and EGH, having the sides AC and CD respectively equal to EG and GH, the angles at D and H right angles, and consequently (I. 32, Cor. 4, and Def. 15) the angles A and E of the same species; to prove that the base AD is equal to the base EH, A ТЕ н the angle A equal to E, and the angle C. equal to G. (Dem.) Since AC is = EG, and CD = GH, therefore AC? — ČD? = EG– GH2; but (I. 47) ADP = AC – CD, and EH= EGʻ– GH; hence ADP = EH?, and therefore AD = EH. Hence the three sides of triangle ACD are respectively equal to the three sides of the triangle EGH, they are therefore every way equal (I. 8); hence, angle A = angle E, and angle C = angle G. Secondly, Given the two triangles ABC and EFG, having the sides AC and CB AD3 E F respectively equal to EG and GF, and the angles BAC and FEG equal and both acute, and the angles ABC and EFG opposite to the other two equal sides, both acute or both obtuse; to prove that the triangles are equal in every other respect. (Const.) Draw CD, GH, perpendicular to the base, or the base produced. (Dem.) Then in the triangles ACD and EGH the right angles at D and H are equal, and also angles at A and E and the sides AC and EG; they are therefore every way equal (I. 26); consequently the sides CD and GH are equal, and AD = EH. Again, the triangles BCD and FGH have two sides BC and CD in the one respectively A B D E F H equal to FG and GH in the other, and the angles at D and H are right angles ; wherefore (by the first case) they are every way equal, and hence DB = HF. Hence the sum of AD and DB in the first two triangles, is equal to the sum of EH and HF, or AB = EF; and their differences in the second pair of triangles are also equal, or AB = EF. Hence the two triangles ABC and EFG have the three sides of the one respectively equal to the three sides of the other, and they are therefore equal in every other respect (I. 8). Hence angle ABC = EFG, and angle ACB = ÉGF. Lastly, when the equal angles BAC and FEG are obtuse. In this case, as in the preceding pair of triangles, the perpendiculars CD and GH fall without the triangles; and since the angles CAD and GEH, adjacent to the equal angles are equal (I, 13, Cor.), and the angles at D and H are right angles; hence the triangles ACD and EGH can be proved equal as in the second case; the tri- D A angles BCD and FGH can be proved equal also, as in that case, and hence AD= EH, and BD = FH; and consequently BD – AD=FH - EH, or AB = EF. Hence (I. 8) the triangles ABC and EFG are equal in every respect. Cor. If any two sides of one right-angled triangle are respec tively equal to two corresponding sides of another, the triangles are every way equal. Bu Egyi EXERCISES. 1. A line * that bisects the vertical angle of an isosceles triangle, also bisects the base perpendicularly. 2. If a line be bisected perpendicularly by another line, every point in the latter is equally distant from the extremities of the former; and any point not situated in the latter is at unequal distances from the extremities of the former. 3. In a given straight line to find a point equally distant from two given points. 4. If any line be drawn through the middle point of the line joining two given points, any two points in the former line that are equidistant from the middle point are also equidistant from the two given points. 5. Of all lines that can be drawn from a given point to a given line, the perpendicular upon it is the least; and of all others, that which is nearer to the perpendicular is less than one more remote; and only two equal lines can be drawn to it from that point, one upon each side of the perpendicular. 276. If from every point of a given line, the lines drawn to each of two given points on opposite sides of the line are equal; prove that the straight line joining the given points will be bisected by the given line at right angles. 7. In the figure to Euc. I. 5, if BG and CF meet in 0, shew that OA bisects the angle BAC. 8. The difference between two sides of a triangle is less than the third side. 9. If two isosceles triangles be constructed on opposite sides of the same base, the line joining their vertices bisects the common base, and each of the vertical angles. 10. Every point in the line that bisects a given angle is equidistant from the sides of the angle. 11. If the alternate extremities of two equal and parallel lines be joined, the connecting lines bisect each other. 3! 12. If the vertical angle of an isosceles triangle be a right angle, each of the angles at the base is half a right angle. 13. If a side of an isosceles triangle be produced beyond the vertex, the exterior angle is double of either of the angles at the base. 14. The middle point of the hypotenuse of a right-angled triangle, is equally distant from each of the three angles. * When the word line is used, straight line is understood, unless otherwise expressed. 15. Of all triangles, having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in that point. 16. If two sides of a triangle be produced, the lines that bisect the two exterior angles, and the angle contained by the two sides produced, pass through the same point. 17. If a right-angled triangle have one of the acute angles double of the other, prove that the hypotenuse is double of the side opposite the least angle. 18. If the exterior angle and one of the opposite interior angles, in one triangle, be respectively double those of another, the remaining opposite interior angle of the former is double that of the latter. · Note.—This proposition is the geometrical principle on which the construction of Hadley's quadrant depends. 19. Through a given point to draw a line such that the segment of it intercepted between two given parallels may be equal to a given line. 20. Through a given point to draw a line that shall be equally inclined to two given lines. 21. If two intersecting lines be respectively parallel, or equally inclined, to other two intersecting lines, the inclination of the former is equal to that of the latter. 22. The sum of two sides of a triangle is greater than twice the line joining the vertex and the middle of the base. 23. Given the sum or difference of the hypotenuse and a side of a right-angled triangle, and also the remaining side, to construct it. 24. Through a given point, between two given lines, to draw a line so that the part of it intercepted between them may be bisected in that point. 25. If two lines bisect the angles at the base of a triangle, the line joining their point of intersection and the vertex bisects the vertical angle. 26. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides, their sum will be equal to the perpendicular from either extremity of the base upon the opposite side. 27. The sum of the perpendiculars drawn from any point within an equilateral triangle on the sides, is equal to the perpendicular from any of the angular points upon the opposite side. 28. If the diagonals of a parallelogram be equal, prove that it is a rectangle. 29. The quadrilateral figure whose diagonals bisect each other is a parallelogram. 30. The sum of the diagonals of any quadrilateral is less than the sum of any four lines that can be drawn to the four corners from any point which is not the intersection of the diagonals. 31. Half the base of a triangle is greater than, equal to, or less than, the line joining the middle of the base and the vertex, according as the vertical angle is obtuse, right, or acute. 32. If two lines bisect perpendicularly two sides of a triangle, the perpendicular, from their point of intersection upon the base, will bisect it. 33. The angle contained by a line drawn from the vertex of a triangle perpendicular to the base, and another bisecting the vertical angle, is equal to half the difference of the angles at the base. 34. To find a point in a given line such, that lines drawn from it to two given points on the same side of the given line, will make equal angles with the given line. 35. Given the sum of the sides of a triangle and the angles at the base, to construct it. 36. If two lines be drawn from the extremities of the base of a triangle to bisect the opposite sides, the line joining their intersection with the vertex, if produced, will bisect the base. 37. If in the figure to (I. 47) DB be produced to meet a perpendicular from F in N, and EC to meet a perpendicular from K in M, and if AL meet BC in 0, prove that FN and KM are each equal to AO, that BN is equal to BO, and CM to CO; and that the two triangles BNF and CMK are together equal to the triangle ABC. 38. If in the figure to (I. 47) DF, GH, and KE be joined, prove that the area of the six-sided figure so formed is equal to twice the square on BC, and four times the triangle ABC. 39. If the two sides that contain the right angle in a rightangled triangle be 160 and 120, prove that the hypotenuse is 200. 40. If the hypotenuse of a right-angled triangle be 169, and one side be 65; prove that the other side is 156. 41. The angles of a quadrilateral are equal to four right angles. 42. Every quadrilateral that has its opposite sides, or opposite angles equal, is a parallelogram. 43. To find a line whose square shall be equal to the difference between two given squares. 44. To find a line whose square shall be equal to the sum of the squares on any number of lines. |