2. Divide a given straight line internally into two segments, such that the rectangle contained by them may be equal to the square on another given straight line. What limits are there to the length of the second straight line? 3. Divide a given straight line externally into two segments, such that the rectangle contained by them may be equal to the square on another given straight line. Are there any limits to the length of the second straight line? 4. Describe a rectangle equal to a given square, and having one of its sides equal to a given straight line. APPENDIX II. PROPOSITION 1. The sum of the squares on two sides of a triangle is double the sum of the squares on half the base and on the median to the base.* Let ABC be a triangle, AD the median to the base BC: it is required to prove AB2 + AC2 = 2 BD2 + 2 AD2. .. AB2 + AC2 = 2 BD2 + 2 AD2. COR. The theorem is true, however near the vertex A may be to the base BC. When A falls on BC, the theorem becomes II. 9; when A falls on BC produced, the theorem becomes II. 10. * Pappus, VII. 122. NOTE.-It may be well to remark that the converse of the theorem, 'If ABC be a triangle, and from the vertex A a straight line AD be drawn to the base BC, so that AB2 + AC2 = 2 BD2 + 2 AD2, then D is the middle point of BC, is not always true. A A B D C' For, let ABC, ABC' be two triangles having AC = AC'. .. B D' C' D D must fall either between B AB2 + AC2 = 2 BD2 + 2 A D2 ; App. II. 1 and we know that D is not the middle point of BC'. .. In the second case, find D' the middle point of BC', and join AD'. and we know that D' is not the middle point of BC. The third case needs no discussion. PROPOSITION 2. App. II. 1 The difference of the squares on two sides of a triangle is double the rectangle contained by the base and the distance of its middle point from the perpendicular on it from the vertex.* A Let ABC be a triangle, D the middle point of the base BC, and AE the perpendicular from A on BC: it is required to prove AB2 - AC2 = 2 BC. DE. * Pappus, VII, 120. For AB2 - AC2 = (BE2 + AE2) − (EC2 + AE2), = I. 47 (BE + EC) (BE – EC), II. 5, 6, Corr. BC.2 DE in fig. 1; or = 2 DE. BC in fig. 2, If the straight line AD be divided internally at any two points C and B, then AC BD + AD · BC = AB. CD.* For ACBD + AD · BC = AC · BD + (BD + AB) · BC, AC. BD + BD · BC + AB · BC, II. 1 = Find the locus of the vertices of all the triangles which have the same base and the sum of the squares of their sides equal to a given square. M Let BC be the given base, M2 the given square. Join AB, AC; bisect BC in D, and join AD. I. 10 * Euler, Novi Comm. Petrop., vol. i. p. 49. Нур. App. II. 1 Then, since A is a point on the locus, AB2 + AC2 = M2. Now M2 is a constant magnitude, and so is BD2, being the square on half the given base; .. M2 - BD2 must be constant ; AD2 must be constant. And since AD2 is constant, AD must be equal to a fixed length; that is, the vertex of any triangle fulfilling the given conditions is always at a constant distance from a fixed point D, the middle of the given base. Hence, the locus required is the Oce of a circle whose centre is the middle point of the base. To determine the locus completely, it would be necessary to find the length of the radius of the circle. This may be left to the reader. PROPOSITION 5. Find the locus of the vertices of all the triangles which have the same base, and the difference of the squares of their sides equal to a given square. Let BC be the given base, M2 the given square. bisect BC in D, and draw AE1 BC or BC produced. Now M2 is a constant magnitude, and so is 2 BC; .. DE must be constant ; I. 10, 12 Нур. App. II. 2 .. a perpendicular drawn to BC from the vertex of any triangle fulfilling the given conditions will cut BC at a fixed point. = If AC2 AB2 M2, the perpendicular from A on BC will cut BC at a point E' on the other side of D, such that DE' = DE. Hence, the locus consists of two straight lines drawn perpendicular to the base and equally distant from the middle point of the base. DEDUCTIONS. 1. If from the vertex of an isosceles triangle a straight line be drawn to cut the base either internally or externally, the difference between the squares on this line and either side is equal to the rectangle contained by the segments of the base. (Pappus, III. 5.) 2. The sum of the squares on the diagonals of a ||m is equal to the sum of the squares on the four sides. 3. The sum of the squares on the diagonals of any quadrilateral is equal to twice the sum of the squares on the straight lines joining the middle points of opposite sides. 4. The sum of the squares on the four sides of any quadrilateral exceeds the sum of the squares on the two diagonals by four times the square on the straight line which joins the middle points of the diagonals. (Euler, Novi Comm. Petrop., i. p. 66.) 5. The centre of a fixed circle is the middle point of the base of a triangle. If the vertex of the triangle be on the ", the sum of the squares on the two sides of the triangle is constant. 6. The centre of a fixed circle is the point of intersection of the diagonals of a ||. Prove that the sum of the squares on the straight lines drawn from any point on the Oce to the four vertices of the m is constant. 7. Two circles are concentric. Prove that the sum of the squares of the distances from any point on the ce of one of the circles to the ends of a diameter of the other is constant. 8. The middle point of the hypotenuse of a right-angled triangle is equidistant from the three vertices. 9. Three times the sum of the squares on the sides of a triangle is equal to four times the sum of the squares on the three medians, or equal to nine times the sum of the squares on the straight lines which join the centroid to the three vertices. 10. If ABCD be a quadrilateral, and P, Q, R, S be the middle points of AB, BC, CD, DA respectively, then 2 PR2 + AB2 + CD2 = 2 QS2 + BC2 + DA2. |