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3. The straight line which joins the middle points of two sides of a triangle is equal to half the third side.

4. Shew that the three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identically equal.

5. Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.

6. Given the three middle points of the sides of a triangle, construct the triangle.

7. AB, AC are two given straight lines, and P is a given point between them ; required to draw through P a straight line terminated by AB, AC, and bisected by P.

8. ABCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: shew that BX and DY trisect the diagonal AC.

9. If the middle points of adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram.

10. Shew that the straight lines which join the middle points of opposite sides of a quadrilateral, bisect one another.

11. The straight line which joins the middle points of the oblique sides of a trapezium, is parallel to the two parallel sides, and passes through the iniddle points of the diagonals.

12. The straight line which joins the middle points of the oblique sides of a trapezium is equal to half the sum of the parallel sides; and the portion intercepted between the diagonals is equal to half the difference of the parallel sides.

DEFINITION. If from the extremities of one straight line perpendiculars are drawn to another, the portion of the latter intercepted between the perpendiculars is said to be the Orthogonal Projection of the first line upon the second.

B

B

Х

Х

P

Y

Thus in the adjoining figures, if from the extremities of the straight line AB the perpendiculars AX, BY are drawn to PQ, then XY is the orthogonal projection of AB on PQ.

13. A given straight line AB is bisected at C; shew that the projections of AC, CB on any other straight line are equal.

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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 X, K.

1. 31. Now AX, CZ, BY are parallel, for they are perp. to PQ ; 1. 28.

.. the figures XH, HY are parms;
:: AH=XZ, and HK=ŻY.

I. 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:
that is, AH=HK;
.: XZ=ZY.

Q.E.D.

Ex. 1,

p. 104.

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 O; 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.

19. 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.

Q 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

Is parallels to BT. It may be shewn by Ex. 14, p. 106, that

T these parallels divide AB into five equal parts.]

R

C 20. If through an angle of a parallelogram any straight line is drawn, the perpendicuiar drawn to it from the opposite angle is equal to the sum or diference of the perpendiculars drawn to it from the two remaining angles, according as the given straight line falls without the parallelogram, 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 froin their point of intersection let fall a perpendicular upon the given straight line. See Ex. 17, p. 106.)

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.

PROBLEMS.

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 Pa 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.

V.

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 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.)

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 4, and AH the bi. sector of the vertical BAC.

Let CLK meet AH at right angles.

(i) Then shall each of the 48 AKC, ACK K
be equal to half the sum of the 28 ABC,
ACB.
In the As AKL, ACL,

B XH
the L KAL=the L CAL,
Because also the L ALK=the L ALC, being rt. 28;

and AL is common to both As;
the L AKL=the L ACL.

Hyp.

1. 26.

Again, the L AKC=the sum of the LS KBC, KCB;

I. 32. : the L ACK=the sum of the L8 KBC, KCB.

To each add the L ACK:
then twice the L ACK=the sum of the L: ABC, ACB;

:: the L ACK=half the sum of the 28 ABC, ACB.

(ii) The L KCB shall be equal to half the difference of the LS ACB, ABC. As before, the L ACK=the sum of the 48 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 Łs ACB, KBC; that is, the L KCB=half the difference of the _s 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.

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