ADDITIONAL PROPOSITIONS. The following Propositions are all important, although they are not included in Euclid's Text. Many of them have been already given as Exercises or Riders ; but they are repeated here in order to bring together the results. As they are important, hints are given in all cases for their solution ; the student will, however, do well to attempt at first unaided the solution of each of these additional propositions, and if he fails he can then make use of the hints supplied. I. — Triangles. 1. If two right-angled triangles have their hypotenuses equal, and one side equal to one side, the triangles shall be equal in all respects. right angles are adjacent; prove that two other sides are in one the two triangles are equal by Prop. 26 ; or, (2) Prove the third sides equal by Prop. 47.] 2. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles opposite to one pair of equal sides DE and BC falls along EF; if ( coincides with F the triangles are mentary.] 3. The straight line which joins the middle points of two sides of a triangle is parallel to the third side, and equal to half of it (see Prop. 39, Ex. 1). 4. The straight line DE drawn through the middle point D of the side AB of the triangle ABC, parallel to the side BC, bisects the third side AC. ADE, CFE equal by Prop. 26.] 5. The medians of a triangle are concurrent, that is, meet at a point, and divide each other in the ratio 2:1. OBD, OCD equal; therefore A0, OD, and AD both bisect ABC; therefore they are identical ; or, (2) Join AO and produce it to meet BC in X and the line BG, which is parallel to OC, in G; CO bisects AG, therefore BOE is parallel to GC and BOCG is a parallelogram, and its diagonals bisect each other.) 6. The straight lines drawn at right angles to the sides of a triangle, from their middle points, are concurrent. OX; prove that OB=0A=0C by Prop. 4, and then by Prop. 8 that OX is at right angles to BC.] 7. The straight lines which bisect the angles of a triangle are concurrent. [Let two bisectors meet in 0 and join OA; draw OL, OM, ON per pendicular to the sides, and prove them equal by Prop. 26 ; prove AN e quals AM by Prop. 47, and that the angle NAO is equal to the angle MAO by Prop. 8.] 8. The perpendiculars from the angles of a triangle on the opposite sides are concurrent. by No. 6 the perpendiculars at A, B, C are concurrent.] 9. To bisect a triangle by a straight line drawn through a given point in one of its sides. (See Note on Additional Proposition, p. 97.) 10. To construct a right-angled triangle, having given the hypotenuse and the sum of the remaining sides. equal to half a right angle, and with centre B and radius equal to required.] 11. To construct a triangle, having given the perimeter and each of the angles at the base. equal to half the given angles respectively ; again make the angles in AB; PQR is the triangle required. Prove by Prop. 32.] 12. To construct a triangle, having given the base, the difference of the angles at the base, and the difference of the remaining sides. of base angles ; with centre A and radius equal to given difference PAB is the triangle required.] 13. To construct a triangle, having its three medians equal to three given straight lines. given medians; finish the construction from proof (2) in No. 5.] II.-Lines. 14. AB is a given straight line bisected at C; show that the projections of AC, CB on any other straight line PQ are equal. in D and E; AD and DE are equal by No. 4 on p. 101 ; and the pro jections are equal by Prop. 34.] 15. To divide a given straight line AB into any number of equal parts. [Draw a line AC making any angle with AB; from AC cut off any part AP, and then parts PQ, QR, RS, ST all equal to AP; join BT, and draw lines through P, Q, R, S parallel to BT; in this case AB will be divided into five equal parts. Prove, by drawing through P, Q, R, S, lines parallel to AB, that the triangles are equal by Prop. 26.] 16. To trisect a given finite straight line. [As in No. 15; or describe an equilateral triangle ABC on AB; bisect the angles CAB, CBA by the lines AO, BO, and draw OD, OE parallel to AC and BC respectively, to meet AB in D, E; prove ODE an equilateral triangle by Props. 29 and 32; and then prove DA DO = DE by Prop. 29.] 17. To find a point in a given straight line such that the sum of its distances from two given fixed points, on the same side of the line, may be the perpendicular to an equal length on the other side of the line ; its distances greater by Props. 4 and 20.] 18. A and B are two fixed points and PQ a given straight line ; find a point X in PQ such that the angle AXP is equal to the angle BXQ. Does it make any difference whether the points A and B are on the same or opposite sides of PQ? [Same as No. 17.] 19. Through a given point P to draw a straight line, such that the part of it intercepted between two given lines AB, AC may be bisected at P. another; EP produced is the line required. Prove by Prop. 26 as in No. 4.] 20. To draw a straight line at right angles to a given straight line from the extremity of it and without producing the given line, [See Ex. 4, p. 69.] 21. To draw straight lines the squares on which shall be equal to two, three, four, etc. times the square on a given straight line. the triangle; the square on the hypotenuse is double the given square. And so on.] 22. Two lines are given in position; find the locus of a point equidistant from them. them. angle so formed and this line is the locus required ; prove by Prop. 26, drawing perpendiculars to the two lines from any point in the bisector.] III.-Parallelograms. 23. The diagonals of a parallelogram bisect each other. [Prove by Props. 29 and 26.] 24. A parallelogram is bisected by any straight line which passes through the middle point of one of its diagonals. [Prove by Props. 29 and 26.] 25. If two equal triangles ABC, ABD are on the same base AB, but on opposite sides of it, the line joining their vertices C, D is bisected by AB. bisected by AB; therefore DC is bisected by AB by No. 4, p. 101.] 26. To inscribe a square in a given triangle. [ABC is the triangle ; draw AD perpendicular to BC; produce BC to E, making CE equal to BD; bisect the angle ADE by DF, meeting AE in F; through F draw FKL parallel to BC meeting AB, AC, AD in L, K, and H respectively ; draw KM, LN perpendicular to BC ; KMNL is the square required. Prove FH equal to HD, and FL equal to HK. This construction assumes that the angles at the base are acute.] MISCELLANEOUS RIDERS ON BOOK I. I.--Depending on the Equality of Triangles. 1. If the straight line joining the vertex of a triangle to the middle point of the base is perpendicular to the base, the triangle is isosceles. 2. If two straight lines bisect each other at right angles, any point in either is equidistant from the extremities of the other line. 3. Find a point which is equidistant from three given points. When is this impossible ? 4. In a given straight line find a point equidistant from two given intersecting straight lines. Is this always possible ? 5. Find a point equidistant from three given straight lines. When is this impossible? 6. Through a given point P draw a straight line making equal angles with two given intersecting straight lines. Can more than one such line be drawn? 7. AB and CD are two lines which meet in an inaccessible point, and AX is drawn at right angles to AB; find a point in AX equidistant from AB and CD. 8. AB and CD are two lines which meet in an inaccessible point, and P is any point between them ; show how to draw a line through P, which, when produced, would pass through the intersection of AB and CD. 9. If it be possible within a quadrilateral ABCD, whose opposite sides are equal, to find a point 0, such that OA, OB, OC, OD are all equal, then AOC, BOD are straight lines and ABCD is equiangular. 10. If the straight line joining the middle points of two opposite sides of a quadrilateral be at right angles to each of these sides, the other two sides of the quadrilateral are equal. 11. If two of the medians of a triangle are equal the triangle is isosceles. 12. Prove by the method of superposition that if two right-angled triangles have their hypotenuses equal and two angles equal, the triangles are equal in all respects. 13. If there be two isosceles triangles on the same base, the straight line which joins their vertices, produced if necessary, bisects the common base, and is perpendicular to it. 14. ABCD is a trapezium, of which the side AB is parallel to the side DC. Show that the area of ABCD is equal to the rea of a parallelogram formed by drawing through M, the middle point of BC, a straight line parallel to AD. 15. If the straight line, joining two opposite angles of a parallelogram, bisects the angles, the parallelogram is a rhombus. 16. If two triangles have two sides of the one equal to two sides of the other, each to each, and the sum of the angles contained by these sides equal to two right angles, the triangles shall be equal in area, |