the same base FB, and between the same parallels FB, GC (th. 23); in like manner, the parallelogram BL is double the triangle ABD, on the same base BD, and between the same parallels BD, AL. And since the doubles of equal things are equal (by ax. 6), therefore the square BG is equal to the parallelogram BL. In like manner, the other square CH is proved equal to the other parallelogram EK. Consequently, the two squares BG and CH together are equal to the two parallelograms BL and EK together, or to the whole square BE; that is, the sum of the two squares on the two less sides is equal to the square on the greatest side. Q. E. D. Corol. 1. Hence the square of either of the two less sides is equal to the difference of the squares of the hypothenuse and the other side (ax. 3); or equal to the rectangle contained by the sum and difference of the said hypothenuse and other side; for (Alg. 13) a2-b2=(a+b) (a-b). Corol. 2. Hence, also, if two right-angled triangles have two sides of the one equal to two corresponding sides of the other, their third sides will also be equal, and the triangles identical. THEOREM XXVII.* In any triangle, the difference of the squares of the two sides is equal to the difference of the squares of the segments of the base, or of the two lines, or distances, included between the extremes of the base and the perpendicular. Let ABC be any triangle having CD perpendicular to AB; then will the difference of the squares of AC, BC be equal to the difference of the squares of AD, Á BD; that is, AC-BC-AD'-BD". *The two following theorems require the aid of the following algebraic formulas : (a + b)2 = a2+2ab+b2=a2 + b2+2ab For, since ACD and BCD are right-angled trianAC' = AD' + CD' gles, and BCBD+CD (by th. 26); .. By subtraction, AC-BC' AD' - BD'. Corol. The rectangle of the sum and difference of the two sides of any triangle is equal to the rectangle. of the sum and difference of the distances between the perpendicular and the two extremes of the base, or equal to the rectangle of the base and the difference or sum of the segments, according as the perpendicular falls within or without the triangle. That is, (AC + BC) . (AC — BC) = (AĎ + BD) . (AD-BD). Or, (AC+ BC). (AC — BC) = AB (AD — BD) in the 2d fig. And, (AC + BC) . (AC — BC) = AB. (AD + BD) in the 1st fig. THEOREM XXVIII. In any obtuse-angled triangle, the square of the side subtending the obtuse angle is greater than the sum of the squares of the other two sides, by twice the rectangle of one of the sides containing the obtuse angle and the distance of the perpendicular drawn from the opposite vertex upon this side, from the obtuse angle. Let ABC be a triangle, obtuse-angled at B, and CD perpendicular to AB; then will the square of AC be greater than the squares of AB, BC, by twice the rectangle of AB, BD. That is, AC2=AB2 + BC2 + 2 AB. BD. See the 1st fig. above. = For, AD' (AB + BD)2 = AB' + BD2 + 2AB. BD; adding CD to both members of this equality, AD2 + CD2 = AB2 + BD2 + CD2 + 2AB. BD (ax. 2.) = But AD+CD' AC, and BD + CD2 = BC' (th. 26). Therefore, by substitution in the last equality but one, AC2 = AB' + BC2 + 2AB. BD. Q. E. D. THEOREM XXIX. In any triangle, the square of the side subtending an acute angle is less than the squares of the other two sides, by twice the rectangle of one of the sides containing the acute angle and the distance of the perpendicular upon this side from the acute angle. C C Let ABC be a triangle, having the angle A acute, and CD perpendicular to AB; then will the square of BC be less than the sum of the squares of AB, AC by twice the rectangle of AB, AD; that is, BC2AB+ AC-2 AD. AB. For BD2=(AD~ AB)2 = AD2 + AB- 2AD.AB. And BD' + DC AD2 + DC' + AB- 2AD. AB (ax. 2). = = A B DA D B Therefore BC AC' + AB- 2AD. AB (th. 26). Q. E. D. THEOREM XXX. In any triangle the double of the square of a line drawn from the vertex to the middle of the base, together with double the square of the half base, is equal to the sum of the squares of the other two sides. Let ABC be a triangle, and CD the line drawn from the vertex to the middle of the base AB, bisecting it into the two equal parts AD, DB; then will the sum of the squares of AC, CB be equal to twice the sum of the squares of CD, AD; A or AC2+ CB2 = 2CD + 2AD2. For AC' CD2+ AD2 + 2AD. DE (th. 28). AC2+ BC2CD2 + AD2 + BD2 C DE B = 2CD2 + 2AD2 (by hyp.). Q. E. D. THEOREM XXXI. In any parallelogram the two diagonals bisect each other, and the sum of their squares is equal to the sum of the squares of all the four sides of the parallelogram. Let ABDC be a parallelogram A whose diagonals intersect each other in E; then will AE be equal to ED and BE to EC, and the sum of the squares of AD, BC will be equal to C E B D the sum of the squares of AB, BD, CD, CA; that is, AE = ED, and BE EC, and AD2 + BC AB2 + BD2 + CD2 + CA', = For, in the triangles AEB, DEC, the two lines AD, BC meeting the parallels AB, DC, make the angle BAE equal to the angle CDE, and the angle ABE equal to the angle DCE, and the side AB is equal to the side DC (th. 19); therefore these two triangles are identical, and have their corresponding sides equal (th. 2), viz., AE= DE, and BE EC. Again, since AD is bisected in E, the sum of the squares, CA + CD' 2CE' + 2DE' (th. 30). = In like manner, BA'+ BD2= 2DE+2BE or 2CE'. Therefore, by addition, AB + BD2 + DC2 + CA2= 4CE+4DE' (ax. 2). But because the square of a whole line is equal to 4 times the square of half the line; that is, BC2= 4BE2, and AD'= 4DE'; = Therefore AB2 + BD2 + DC2 + CA2= BC2 + AD2 (ax. 1). Q. E. D. Cor. If AB AC, or the parallelogram be a rhombus, then the triangles AEB, AEC will be mutually equilateral, and, consequently (th. 5), the angle BEA of the one will be equal to the angle AEC of the other. Hence (def. 12) the diagonals of a rhombus intersect at right angles. *This may be seen from the accompanying diagram, or, algebraically, from considering that the square of ¦ļa is ļa2. EXERCISES. 1. To construct an isosceles triangle with a given base and given vertical angle. 2. Prove that every point of the bisectrix of a given angle is equally distant from the sides. 3. Two angles of a triangle being given, to find the third. 4. To construct an isosceles triangle so that the vertex shall fall at a given point, and the base fall in a given line. 5. An isosceles triangle so that the base shall be a given line and the vertical angle a right angle. 6. With two angles and a side opposite one of them, to construct a triangle. 7. To construct a triangle when the base, the angle opposite, and the sum of this and one of the other two angles are given. 8. The same, except the difference instead of the sum given. 9. To construct a quadrilateral when the four sides and one angle are given. 10. When three of the sides and the two angles included between them are given. 11. When two sides and the included angle and two other angles. 12. To construct a parallelogram with two adjacent sides and the diagonal given. 13. To construct a parallelogram with given base, altitude, and diagonal. 14. With two adjacent sides and the altitude. 15. To make a hexagon equal in all respects to a given irregular hexagon. 16. To construct a triangle with the angles at the base and the altitude given. 17. With the vertical angle, one of its sides and the altitude given. 18. With the base, altitude, and one of the angles at the base given. 19. To construct a trapezoid when three sides and the angle contained between two of them are given. 20. A line and two points without it being given, to find a point in the line equidistant from the two given points.* * Geometric Analysis.--The best method for discovering the solution of problems is what is termed the analytic. This consists in supposing the problem solved, making the diagram accordingly, and then, by examination of the required and given parts of the diagram in their relations to one another, considering what known theorems of geometry connect them together. This is a sort of going back from the result sought by a chain of relations--depending upon known |