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13. A given straight line AB is bisected at C; shew that the projections of AC, CB on any other straight line are equal.
Let XZ, ZY be the projections of AC, CB on any straight line PQ.
Then XZ and ZY shall be equal. Through A draw a straight line parallel to PQ, meeting CZ, BY or these lines produced in H, K.
.. the figures XH, HY are parms ;
1. 34. But through C, the middle point of AB, a side of the A ABK, CH has been drawn parallel to the side BK; :: CH bisects AK:
104. that is, AH=HK; .: XZ=ZY.
14. If three parallel straight lines make equal intercepts on a fourth straight line which meets them, they will also make equal intercepts on any other straight line which meets them.
15. Equal and parallel straight lines have equal projections on any other straight line.
16. AB is a given straight line bisected at 0; and AX, BY are perpendiculars drawn from A and B on any other straight line: shew that OX is equal to OY.
17. AB is a given straight line bisected at O: and AX, BY and OZ are perpendiculars drawn to any straight line PQ, which does not pass between A and B : shew that OZ is equal to half the sum of AX, BY.
[OZ is said to be the Arithmetic Mean between AX and BY.]
18. AB is a given straight line bisected at 0; and through A, B and O parallel straight lines are drawn to meet a given straight line PQ in X, Y, Z: shew that OZ is equal to half the sum, or half the difference of AX and BY, according as A and B lie on the same side or on opposite sides of PQ.
To divide a given finite straight line into any number of equal parts.
[For example: required to divide the straight line AB into five equal parts.
From A draw AC, a straight line of unlimited length, making any angle with AB.
In AC take any point P; and by marking off successive parts PQ, QR, RS, SŤ each equal to AP, make AT to contain AP five times.
Join BT; and through P, Q, R, S draw parallels to BT. It may be shewn by Ex. 14, p. 106, that
T these parallels divide AB into five equal parts.]
20. If through an angle of a parallelogram any straight line is drawn, the perpendicular drawn to it from the opposite angle is equal to the sum or difference of the perpendiculars drawn to it from the two remaining angles, according as the given straight line falls without the parallelogran, or intersects it.
[Through the opposite angle draw a straight line parallel to the given straight line, so as to meet the perpendicular from one of the remaining angles, produced if necessary; then apply 1. 34, 1. 26. Or proceed as in the following example.]
21. From the angular points of a parallelogram perpendiculars are drawn to any straight line which is without the parallelogram : shew that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair.
[Draw the diagonals, and from their point of intersection let fall a perpendicular upon the given straight line. See Ex. 17,
22. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal sides is equal to the perpendicular drawn from either extremity of the base to the opposite side.
[It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken.]
23. In the base produced of an isosceles triangle any point is taken: shew that the difference of its perpendicular distances from the equal sides is constant.
24. The sum of the perpendiculars drawn from any point within an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore constant.
25. Draw a straight line through a given point, so that the part of it intercepted between two given parallel straight lines may be of given length. When does this problem admit of two solutions, when of only one, and when is it impossible ?
26. Draw a straight line parallel to a given straight line, so that the part intercepted between two other given straight lines may be of given length.
27. Draw a straight line equally inclined to two given straight lines that meet, so that the part intercepted between them may be of given length.
28. AB, AC are two given straight lines, and P is a given point without the angle contained by them. It is required to draw through P a straight line to meet the given lines, so that the part intercepted between them may be equal to the part between P and the nearer line.
MISCELLANEOUS THEOREMS AND EXAMPLES.
Chiefly on 1. 32. 1. A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that AD is equal to BA; if DC is drawn, shew that BCD is a right angle.
2. The straight line joining the middle point of the hypotenuse of a right-angled triangle to the right angle is equal to half the hypotenuse.
3. From the extremities of the base of a triangle perpendiculars are drawn to the opposite sides (produced if necessary); shew that the straight lines which join the middle point of the base to the feet of the perpendiculars are equal.
4. In a triangle ABC, AD is drawn perpendicular to BC; and X, Y, Z are the middle points of the sides BC, CA, AB respectively : shew that each of the angles ZXY, ZDY is equal to the angle BAC.
5. In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypotenuse, the two triangles thus formed are equiangular to one another.
6. In a right-angled triangle two straight lines are drawn from the right angle, one bisecting the hypotenuse, the other perpendicular to it : shew that they contain an angle equal to the difference of the two acute angles of the triangle. (See above, Ex. 2 and Ex. 5.)
V 7. In a triangle if a perpendicular is drawn from one extremity
of the base to the bisector of the vertical angle, (i) it will make with either of the sides containing the vertical angle an angle equal to half the sum of the angles at the base ; (ii) it will make with the base an angle equal to half the difference of the angles at the base.
Let ABC be the given A, and AH the bi. sector of the vertical L BAC.
Let CLK meet AH at right angles.
(i) Then shall each of the L8 AKC, ACK
and AL is common to both As ;
Again, the L AKC=the sum of the 48 KBC, KCB;
1. 32. :: the L ACK=the sum of the 48 KBC, KCB.
To each add the L ACK:
: the ACK=half the sum of the Lo ABC, ACB.
(ii) The L KCB shall be equal to half the difference of the 28 ACB, ABC. As before, the L ACK=the sum of the L8 KBC, KCB.
To each of these add the L KCB : then the L ACB=the L KBC together with twice the L KCB. :: twice the L KCB=the difference of the 4* ACB, KBC; that is, the L KCB=half the difference of the 25 ACB, ABC.
COROLLARY. If X is the middle point of the base, and XL is joined, it may be shewn by Ex. 3, p. 105, that XL is half BK; that is, that XL is half the difference of the sides AB, AC.
8. In any triangle the angle contained by the bisector of the vertical angle and the perpendicular from the vertex to the base is equal to half the difference of the angles at the base.
[See Ex. 3, p. 65.] 9. In a triangle ABC the side AC is produced to D, and the angles BAC, BCD are bisected by straight lines which meet at F; shew that they contain an angle equal to half the angle at B.
10. If in a right-angled triangle one of the acute angles is double of the other, shew that the hypotenuse is double of the shorter side.
11. If in a diagonal of a parallelogram any two points equidistant froin its extremities are joined to the opposite angles, the
figure thus formed will be also a parallelogram. 1
12. ABC is a given equilateral triangle, and in the sides BC, CA, AB the points X, Y, Z are taken respectively, so that BX, CY and AŽ are all equal. AX, BY, CZ are now drawn, intersecting in P, Q, R: shew that the triangle PQR is equilateral.
13. If in the sides AB, BC, CD, DA of a parallelogram ABCD four points P, Q, R, S are taken in order, one in each side, so that AP, BQ, CR, DS are all equal; shew that the figure PQRS is a parallelogram.
14. In the figure of 1. 1, if the circles intersect at F, and if CA and CB are produced to meet the circles in P and Q respectively; shew that the points P, F, Q are in the same straight line; and shew also that the triangle CPQ is equilateral. [Problems marked (*) admit in general of more than one solution.]
15. Through two given points draw two straight lines forming with a straight line given in position, an equilateral triangle.
*16. From a given point it is required to draw to two parallel straight lines two equal straight lines at right angles to one another.
*17. Three given straight lines meet at a point; draw another straight line so that the two portions of it intercepted between the given lines may be equal to one another.
18. From a given point draw three straight lines of given lengths, so that their extremities may be in the same straight line, and intercept equal distances on that line.
[See Fig. to 1. 16.] 19. Use the properties of the equilateral triangle to trisect a given finite straight line.
20. In a given triangle inscribe a rhombus, having one of its angles coinciding with an angle of the triangle.
ON THE CONCURRENCE OF STRAIGHT LINES IN A
DEFINITIONS. (i) Three or more straight lines are said to be concurrent when they meet in one point.
(ii) Three or more points are said to be collinear when they lie upon one straight line.
Obs. We here give some propositions relating to the concurrence of certain groups of straight lines drawn in a triangle : the importance of these theorems will be more fully appreciated when the student is familiar with Books III. and iv.