19. If perpendiculars be drawn from the extremities of a diameter to any chord in the circle, they will cut off equal segments. 20. If, from the extremities of any chord in a circle, straight lines be drawn to any point in the diameter to which it is parallel, the sum of their squares will be equal to the sum of the squares of the segments of the diameter. 21. If, from two angular points of any triangle, straight lines be drawn, to bisect the opposite sides, they will divide each other into segments, having the ratio of two to one, and, if a line be drawn through the third angle and their point of intersection, it will bisect the third side, and divide the triangle into six equal triangles. 22. If two triangles have two angles equal, and other two angles together equal to two right angles, the sides about the remaining angles will be proportional. 23. If a circle be described on the radius of another circle, any straight line drawn from the point where they meet to the outer circumference is bisected by the interior one. 24. The rectangle contained by two sides of a triangle, is equal to the rectangle contained by the perpendicular from the contained angle upon the third side, and the diameter of the circumscribing circle. 25. If a straight line be drawn bisecting the vertical angle of a triangle, the rectangle contained by th two sides will be equal to the square of the bisecting line, together with the rectangle contained by the segments of the base. 26. A straight line drawn from the vertex of a triangle to meet the base, divides a parallel to the base in the same ratio as the base. 27. If the three sides of one triangle be perpendicular to the three sides of another, the two triangles are similar. 28. If from any point in the diameter of a circle produced, a tangent be drawn, a perpendicular from the point of contact to the diameter will divide it into segments, which have the same ratio which the distances of the point without the circle from each extremity of the diameter have to each other. 29. A straight line drawn from the vertex of an equilateral triangle, inscribed in a circle to any point in the opposite circumference, is equal to the two lines together which are drawn from the extremities of the base to the same point. 30. If the interior and exterior vertical angles of a triangle be bisected by straight lines, which cut the base, and the base produced, they will divide it into three segments, such that the rectangle contained by the whole base thus produced, and the middle segment, shall be equal to the rectangle contained by the two extreme segments. (A line divided in this manner is said to be divided harmonically.) 31. The side of a regular decagon inscribed in a circle, is equal to the greater segment of the radius divided medially; the side of a regular hexagon is equal to the radius; and the side square of a regular pentagon is equal to the side square of a regular hexagon, tugether with the side square of a regular decagon. L B A To divide a given angle ABC into two equal parts. From the centre B with any radius, describe the arc AC, and from the centres A and C with the same radius, describe arcs intersecting in D, and join BD; the angle ABC will be bisected by the (BD. PROBLEM III. To trisect a right angle ABC'; that is, to divide it into three equal parts. From the centre B with any radius, describe the arc AC; and from the centre C with the same radius, cut the arc in D, and from the centre A with the same radius, cut the arc in E, and join BD and BE, and they will trisect the angle as required. PROBLEM IV. To erect a perpendicular from a given point A, in a given line AC. CASE I. When the point is near the middle of the line. On each side of the point A take any two equal distances, Am, An. From the centres m, n, with any radius greater than Am or An, describe two arcs intersect Bm A in А ing in D. Through A and D draw the straight line AD, and it will be the perpendicular required. CASE II. When the point is near the end of the line. From the centre A with any radius, describe an arc mnr. From the centre m with the same radius, turn the compasses twice over on the arc at n and r. Again, from the centres n and r with the same Bradius, describe arcs intersecting in D. Then draw AD, and it will be the perpendicular required. Another Method. From any point n as a centre, with a radius =nA, describe an arc, not less than a semicircle, cutting the line in m and A. Through m and n draw the diameter mnr, cutting the arc in r, and B join Ar, and it will be the perpendicular required. Cor. If the point from which the perpendicular is to be drawn were the extremity of the line C, it would only be necessary to take nC as a radius instead of nA, and the other parts of the construction would be the same. m Another Method. From any scale of equal parts take a distance equal to 3 divisions; then from the centre A, with a radius = 4 divisions, describe an arc, and from the centre m, with a radius = 5 divisions, describe another arc, cutting B the former in D, and join DA, and it will be the perpendicular required. m A Another Method. With a marquois square, which is a right-angled triangle cut out of ivory or wood, apply the right angle to the point A, and make one side coincide with AB, a line drawn along the other side will be the perpendicular required. PROBLEM V. From a given point C, without a straight line AB, to let fall a perpendicular. D B m Case I. When the point is nearly opposite to the middle of the line. From the centre C describe an arc, cutting the straight line AB in m and n. From the centres m and n describe arcs, with any radius greater than mn, inter- Asecting in s. A straight line drawn through the points C and 8 will be perpendicular to AB. Or, apply one of the sides containing the right angle of a marquois square to AB, and slip it along the line, till the point C coincide with the other side; then a line drawn along this side will be perpendicular to AB. CASE II. When the point is nearly opposite to the end of the line. From C draw any line Cm, and bisect it in n. From the centre n, with the radius Cn, describe the semicircle CDm, cutting the line AB in D; join CD, and it will be the perpendicular required. n A m B Another Method. B D From A, or any point in AB as a centre, describe an arc through the point C. From any other point m, describe another arc through C, Acutting the former in n. Through C and n draw the line CDn; and CD will be perpendicular to AB. N.B.—This can also be done as described in Case I., by the marquois square. n PROBLEM VI. At a given point E in the line ED, to make an angle equal to a given angle ABC. From the point B with any radius, describe the arc mn, cutting AB and CB in m and B. n. From the centre E with the radius Bm, describe the arc sr. Take the dis n tance mn, and lay it from r to s, and through Es draw the straight line EsF; then the angle FED will be = the LABC. If the angle be given in degrees. Draw a straight line as ED, and using a scale of chords from the centre E, with a radius = the chord of 60°, describe an arc rs; then take off the chord of the number of degrees required from the scale, and from the centre r, with the chord of the given angle as a radius, cut the arc in s, and through the points E and s draw the EF, and the _FED will contain the given number of degrees. If the angle be greater than 90°, it can be divided into two parts, and one part laid off, and then the other part, from its extremity; it is generally most convenient to lay off the chord of 60°, and then the chord of the excess above 600. PROBLEM VII. To draw a line parallel to a given line AB. CASE I. When the parallel line is to be at a given disance C, from the given line. From any two points m and n n the line AB, with a radius qual to C, describe the arcs r Am od s. Draw DE to touch these rcs without cutting them, and will be the parallel required. CASE II. When the parallel line is to pass through a ven point C. From any point m in the line B, with the radius mC, describe le arc Cn. From the centre C ith the radius Cm, describe the Take the arc Cn in the impasses, and apply it from m to 0. Through C and o aw DE, and it will be the parallel required. V.B.—This problem is more easily effected with a paHel ruler. PROBLEM VIII. D A B Сто. |