Similarly, BC is equal to BE. Also, the angle HBC is equal to the angle ABE, the sum of the angle ABC and a right angle. because the sides HB, BC and their included angle HBC, in the triangle HBC, are equal respectively to the sides AB, BE and their included angle ABE, in the triangle ABE, therefore the triangles HBC, ABE are equal in area. [Prop. 4] Again, because AKHB and CHB are respectively a parallelogram and a triangle on the same base HB and between the same parallels KC, HB, therefore the area of AKHB is double that of CHB. [Prop. 41] Similarly, the area of the parallelogram BL is double that of the triangle ABE. so that, the area of AKHB is equal to that of BL. the area of ACFG is equal to that of the parallelogram CL. So that the whole area of BEDC is equal to the sum of the areas of CFGA, AKHB. Wherefore, in a right-angled triangle, etc. Q.E.D. EXAMPLES XLIX. 1. The sum of the areas of the four squares on the sides of a rectangle is equal to the sum of the areas of the two squares on its diagonals. 2. The square on the diagonal of a square is equal in area to twice the square. 3. The square on the hypotenuse of an isosceles right-angled triangle is equal in area to four times the square on the perpendicular from the right angle to the hypotenuse. 4. Prove from Question 3 that the square on a line is equal in area to four times the square on half the line. [Prove it also independently.] 5. The square on the perpendicular from an angular point of an equilateral on the opposite side is equal in area to three times the square on half the side. 6. Prove that if the squares on two lines are of equal area, the lines must be equal. [Use the method of superposition.] 7. If two right-angled triangles have, one side and the hypotenuse in one triangle, equal to the corresponding sides of the other, prove by Prop. 47, that their third sides are equal. 8. In any triangle ABC, if D be the point in which the perpendicular from A cuts BC, then the difference of the areas of the squares on AB, AC is equal to the difference of the areas of the squares on DB, DC. 9. A line CD is drawn perpendicular to the line joining two given points A, B; prove that the difference (of the areas) of the squares on the lines drawn from any point P on CD to A and B respectively is constant. 10. BAC is a right angle, and D, E are points in AC, AB respectively; prove that the sum of the squares on DB, EC is equal (in area) to the sum of the squares on BC, DE. 11. ABCD is a rectangle and O is any point in its plane; prove that squares on OA, OC= in area squares on OB, OD. 12. When the diagonals of a quadrilateral are perpendicular to each other then the sum of the squares on two opposite sides is equal (in area) to the sum of the squares on the other two sides. 13. Shew how to describe a square equal in area to the sum of two given squares. 14. ABC is a triangle and P any point within it, PD, PE, PF are perpendiculars on the sides BC, CA, AB; prove that the sum of the squares on BD, CE, AF is equal in area to the sum of the squares on CD, BF, AE. 15. In the figure of the 47th Proposition, prove that (i) KB, GC are parallel, (ii) HAF is a straight line, (iii) FAH passes through the point of intersection of GK, CB. FG. (iv) LA passes through the point of intersection of HK, (v) AE, HC are at right angles, (vi) the angle HBE is the supplement of ABC, (vii) the triangles HBE, FCD, KAG, ABC are equal in area. 16. In any triangle the sum of the squares on the sides containing an acute angle is greater (in area) than the square on the side opposite to that acute angle. 17. In any obtuse-angled triangle the sum of the squares on the sides containing the obtuse angle is greater (in area) than the square on the side opposite to the obtuse angle. 18. When the diagonals of a quadrilateral are not at right angles then the sum of the areas of the squares on one pair of opposite sides is greater than the sum of the areas of the squares on the other pair of opposite sides. 146. Proposition 48. When in a triangle, the area of the square on one of the sides is equal to the sum of the areas of the two squares on the other sides, then the angle between those two sides is a right angle. Let ABC represent a triangle in which the area of the square on BC is equal to the area of the two squares on AB, AC; that the angle BAC is a right angle. B Let AD be at right angles to CA, and let AD be equal to AB. Now, because AD is equal to AB, therefore the sum of the areas of the squares on DA, AC is equal to the sum of the areas of the squares on BA, AC. Again, because CAD is a right angle, therefore the area of the square on DC is equal to the sum of the areas of the squares on DA, AC. [Prop. 47] And it is given that the area of the square on CB is equal to the sum of the areas of the squares on BA, AC; so that, the square on DC is equal to the square on BC. Whence it follows that the line DC is equal to BC. Consider the triangles BAC, DAC; because the sides BA, AC, CB are equal respectively to the sides DA, AC, CD, therefore the triangles BAC, DAC are equal in all respects; [Prop. 8] so that, the angle BAC is equal to and DAC is a right angle; so that BAC is a right angle. Wherefore, when the area of the square on one side, etc. Q.E.D. EXAMPLES L. 1. ABC, DBC are two right-angled triangles having a common hypotenuse BC; prove that if AB is greater than DB then AC is less than DC. 2. If the difference of the areas of the squares on two sides of a triangle is equal to the area of the square on the third side the triangle is right-angled. 3. Part of the locus of a point P which is such that the difference of (the areas of) the squares described on the distances of P from two given points A, B is constant, is a straight line, perpendicular to AB. [See Question 9, Examples XLIX.] 4. Two isosceles triangles have their equal sides of the same length; prove that if they are of the same altitude they are equal in all respects. 5. Two isosceles triangles have their equal sides of the same length; prove that the triangle which has the greater altitude has the shorter base. 6. If the sum of the (areas of the) squares on two opposite sides of a quadrilateral is equal to the sum of (the areas of) the squares on the other two sides, the diagonals of the quadrilateral are at right angles. [See Question 12, Examples XLIX.] 7. If the squares on two sides of a triangle together are greater (in area) than the square on the third side then the angle contained by the two sides is less than a right angle. |