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VII.

ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN

PARTS.

CLE

Obs. No general rules can be laid down for the solution of problems in this section ; but in a few typical cases we give constructions, which the student will find little difficulty in adapting to other questions of the same class.

1. Construct a right-angled triangle, having given the hypotenuse and the sum of the remaining sides.

[It is required to construct a rt.angled A, having its hypotenuse equal to the given straight line K, and the sum of its remaining sides equal to AB. From A draw AE making with BA

K an equal to half a rt. L. From centre B, with radius equal to K, describe a circle cutting AE in the points C, C.

A D D B From C and C draw perps CD, C'D' to AB; and join CB, C'B. Then either of the As CÔB, C'D'B will satisfy the given conditions.

NOTE. If the given hypotenuse K be greater than the perpendicular drawn from B to AE, there will be two solutions. If the line K be equal to this perpendicular, there will be one solution ; but if less, the problem is impossible.]

2. Construct a right-angled triangle, having given the hypotenuse and the difference of the remaining sides.

3. Construct an isosceles right-angled triangle, having given the sum of the hypotenuse and one side.

4. Construct a triangle, having given the perimeter and the angles at the base.

Х

Y

A

R

B

[Let AB be the perimeter of the required A, and X and Y the 28 at the base.

From A draw AP, making the L BAP equal to half the L X.
From B draw BP, making the L ABP equal to half the . Y.
From P draw PQ, making the L APQ equal to the L BAP.
From P draw PR, making the L BPR equal to tie L ABP.

Then shall PQR be the required A.]

5. Construct a right-angled triangle, having given the perimeter and one acute angle.

6. Construct an isosceles triangle of given altitude, so that its base may be in a given straight line, and its two equal sides may pass through two fixed points.

(See Ex. 7, p. 55.] 7. Construct an equilateral triangle, having given the length of the perpendicular drawn from one of the vertices to the opposite side.

8. Construct an isosceles triangle, having given the base, and the difference of one of the remaining sides and the perpendicular drawn from the vertex to the base.

[See Ex. 1, p. 96.] 9. Construct a triangle, having given the base, one of the angles at the base, and the sum of the remaining sides.

10. Construct a triangle, having given the base, one of the angles at the base, and the difference of the remaining sides. [Two cases arise, according as the given angle is adjacent to the greater side or the less.]

11. Construct a triangle, having given the base, the difference of the angles at the base, and the difference of the remaining sides.

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[Let AB be the given base, X the difference of the 28 at the base, and K the difference of the remaining sides.

Draw BE, making the L ABE equal to half the 2 X.
From centre A, with radius equal to K, describe a circle cutting
BE in D and D'. Let D be the point of intersection nearer to B.

Join AD and produce it to C.
Draw BC, making the LDBC equal to the L BDC.

Then shall CAB be the required. Ex. 7, p. 109. NOTE. This problem is possible only when the given difference K is greater than the perpendicular drawn from A to BE.]

12. Construct a triangle, having given the base, the difference of the angles at the base, and the sum of the remaining sides.

13. Construct a triangle, having given the perpendicular from the vertex on the base, and the difference between each side and the adjacent segment of the base.

14. Construct a triangle, having given two sides and the median which bisects the remaining side.

[See Ex. 18, p. 110.] 15. Construct a triangle, having given one side, and the medians which bisect the two remaining sides.

[See Fig. to Ex. 4, p. 113.

Let BC be the given side. Take two-thirds of each of the given medians; hence construct the triangle BOC. The rest of the construction follows easily.]

16. Construct a triangle, having given its three medians. [See Fig. to Ex. 4, p. 113.

Take two-thirds of each of the given medians, and construct the triangle OKC. The rest of the construction follows easily.]

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See Propositions 35—48. Obs. It must be understood that throughout this section the word equal as applied to rectilineal figures will be used as denoting equality of area unless otherwise stated,

1. Shew that a parallelogram is bisected by any straight line which passes through the middle point of one of its diagonals.

[1. 29, 26.] 2. Bisect a parallelogram by a straight line drawn through a given point.

3. Bisect a parallelogram by a straight line drawn perpendicular to one of its sides.

4. Bisect a parallelogram by a straight line drawn parallel to a given straight line.

5. ABCD is a trapezium in which the side AB is parallel to DC. Shew that its area is equal to the area of a parallelogram formed by drawing through X, the middle point of BC, a straight line parallel to AD, meeting DC, or DC produced.

[1. 29, 26.] 6. A trapezium is equal to a parallelogram whose base is half the sum of the parallel sides of the given figure, and whose altitude is equal to the perpendicular distance between them.

7. ABCD is a trapezium in which the side AB is parallel to DC; shew that it is double of the triangle

formed by joining the extremities of AD to X, the middle point of BC.

8. Shew that a trapezium is bisected by the straight line which joins the middle points of its parallel sides.

[1. 38.] Obs. In the following group of Exercises the proofs depend chiefly on Propositions 37 and 38, and the two converse theorems.

9. If two straight lines AB, CD intersect at X, and if the straight lines AC and BD, which join their extremities are parallel, shew that the triangle AXD is equal to the triangle BXC.

10. If two straight lines AB, CD intersect at X, so that the triangle AXD is equal to the triangle XCB, then AC and BD are parallel

11. ABCD is a parallelogram, and X any point in the diagonal AC produced ; shew that the triangles XBC, XDC are equal. [See Ex. 13, p. 70.]

12. ABC is a triangle, and R, Q the middle points of the sides AB, AC; shew that if BQ and CR intersect in X, the triangle BXC is equal to the quadrilateral AQXR. [See Ex. 5, p. 79.]

13. If the middle points of the sides of a quadrilateral be joined in order, the parallelogram so formed (see Ex. 9, p. 105] is equal to half the given figure.

14. Two triangles of equal area stand on the same base but on opposite sides of it: shew that the straight line joining their vertices is bisected by the base, or by the base produced.

15. The straight line which joins the middle points of the diagonals of a trapezium is parallel to each of the two parallel sides.

16. (i) A triangle is equal to the sum or difference of two triangles on the same base (or on equal bases), if the altitude of the first is equal to the sum or difference of the altitudes of the others.

(ii) A triangle is equal to the sum or difference of two triangles of the same altitude, if the base of the first is equal to the sum or difference of the bases of the others.

Similar statements hold good of parallelograms.

17. ABCD is a parallelogram, and O is any point outside it; shew that the sum or difference of the triangles OAB, OCD is equal to half the parallelogram. Distinguish between the two cases.

Obs. On the following proposition depends an important theorem in Mechanics : we give a proof of the first case, leaving the second case to be deduced by a similar method.

18. (i) ABCD is a parallelogram, and O is any point without the angle BAD and its opposite vertical angle; shew that the triangle OAC is equal to the sum of the triangles OAD, OAB.

(ii) If O is within the angle BAD or its opposite vertical angle, the triangle OAC is equal to the difference of the triangles OAD, OĂB.

CASE I. If O is without the L DAB and its opp. vert. L, then OA is without the parin ABCD: therefore the perp. drawn from C to OA is equal to the sum of the perps drawn from B and D to OA. [See Ex. 20, p. 107.]

Now the As OAC, OAD, OAB are
upon the saine base OA;
and the altitude of the A OAC with

respect to this base has been shewn to A
be equal to the sum of the altitudes of
the As OAD, OAB.

Therefore the A OAC is equal to the sum the As OAD, OAB. [See Ex. 16, p. 118.]

Q.E. D.

19. ABCD is a parallelogram, and through O, any point within it, straight lines are drawn parallel to the sides of the parallelogram; shew that the difference of the parallelograms DO, BO is double of the triangle AOC. [See preceding theorem (ii).]

20. The area of a quadrilateral is equal to the area of a triangle having two of its sides equal to the diagonals of the given figure, and the included angle equal to either of the angles between the diagonals.

21. ABC is a triangle, and D is any point in AB; it is required to draw through D a straight line DE to meet BC produced in E, so that the triangle DBE may be equal to the triangle ABC.

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1. 31.

[Join DC. Through A draw AE parallel to DC.

Join DE.
The A EBD shall be equal to the A ABC.]

I. 37.

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