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Let L be the given line and P be the point.

E

D

cut L in two points A and D.

Then PD is the required.

Constr.-Draw any line through P meeting L at some point A. Bisect AP in C (119°, Cor. 1), and with C as centre and CP as radius describe a circle.

Proof. PDA is the angle in a semicircle,

LPDA is a.

A

If PA is not to L, the ○ will

(106°, Cor. 4) Cor. Let D be a given point in L. With any centre C and CD as radius describe a circle cutting L again in some point A. Draw the radius ACP, and join D and P. Then DP is to L.

S

.. the construction draws a to L at a given point in L. (Compare 119°, Cor. 2)

Cor. 2. Let L be a given line and C a given point.
To draw through C a line parallel to L.

With C as the centre of a circle, construct a figure as given. Bisect PD in E (119°, Cor. 1). Then CE is || to L. For C and E are the middle points of two sides of a triangle of which L is the base. (84°, Cor. 2)

121°. The Square.—The square consists of two rules with their edges fixed permanently at right angles, or of a triangular plate of wood or metal having two of its edges at right angles.

To test a square.

Draw a line AB and place the square as at S, so that one edge coincides with the line, and along the other edge draw the line CD.

B

S'

Next place the square in the position S'.

If the edges can

1

now be made to coincide with the two lines the square is true.

This test depends upon the fact that a right angle is onehalf a straight angle.

The square is employed practically for drawing a line to another line.

OUT

Cor. 1. The square is employed to draw a series of parallel lines, as in the figure.

Cor. 2. To draw the bisectors of an angle by means of the square.

Let AOB be the given angle. Take OA=OB, and at A and B draw perpendiculars to OA and OB.

Since AOB is not a straight angle, these perpendiculars meet at some point C. (79°, Cor.)

Then OC is the internal bisector of LAOB.

For the tri

angles AOC and BOC are evidently congruent.

LAOC=LBOC.

The line drawn through O to OC is the external bisector.

122°. Problem.—Through a given point in a line to draw a line which shall make a given angle with that line.

B

Let P be the given point in the line L, and let X be the given angle.

Constr.- From any point B in the arm OB draw a 1 to the arm OA.

A

A

(120°)

Make PA'=OA, and at A' draw the ▲ A'B' making A'B'=AB. PB' is the line required.

Proof. The triangles OBA and PB'A' are evidently congruent, and .. LBOA=X=LB'PA'.

Cor. Since PA' might have been taken to the left of P, the problem admits of two solutions. When the angle X is a right angle the two solutions become one.

123°. The Protractor.-This instrument has different forms

depending upon the accuracy required of it. It usually consists of a semicircle of metal or ivory divided into degrees, etc. (41°). The point C is the centre. By placing the straight edge of the instrument in coincidence with a given line AB so that the centre falls at a given point C, we can set off any angle given in degrees, etc., along the arc as at D. Then the line CD passes through C and makes a given angle with AB.

B

124°. Problem.-Given the sides of a triangle to construct it. Constr.-Place the three sides of the triangle in line, as AB, BC, CD.

E

With centre C and radius CD
D describe a circle, and with centre

A

B

с

B and radius BA describe a circle.

Let E be one point of intersection of these circles.
Then ABEC is the triangle required.

Proof.-BE-BA and CE=CD.

Since the circles intersect in another point E', a second triangle is formed. But the two triangles being congruent are virtually the same triangle.

Cor. I. When AB=BC=CA the triangle is equilateral.
(53°, Def. 2)

In this case the circle AE passes through C and the circle DE through B, so that B and C become the centres and BC a common radius.

Cor. 2. When BC is equal to the sum or difference of AB and CD the circles touch (113°, Def.) and the triangle takes the limiting form and becomes a line.

When BC is greater than the sum or less than the differ

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ence of AB and CD the circles do not meet (113°, Def.) and no triangle is possible.

Therefore that three line-segments may form a triangle, each one must be less than the sum and greater than the difference of the other two.

125°. The solution of a problem is sometimes best effected by supposing the construction made, and then by reasoning backwards from the completed figure to some relation amongst the given parts by means of which we can make the construction.

This is analogous to the process employed for the solution of equations in Algebra, and a more detailed reference will be made to it at a future stage.

The next three problems furnish examples

126°. Problem.-To construct a triangle when two sides and the median to the third side are given.

Let a and b be two sides and n the median to the third side.

Suppose ACB is the required triangle having CD as the given median.

B

D

F

Ά

a

b

n

By completing the ACBC' and joining DC', we have DC'

equal to CD and in the same line, and BC'=AC (81°); and the triangle CC'B_has_CC'=2n, CB=a, and BC'=AC=b, and is constructed by 124°.

Thence the triangle ACB is readily constructed.

Cor. Since CC' is twice the given median, and since the possibility of the triangle ACB depends upon that of CC'B, therefore a median of a triangle is less than one-half the sum, and greater than one-half the difference of the conterminous sides. (124°, Cor. 2)

127°. Problem.-To trisect a given line-segment, i.e., to divide it into three equal parts.

A

D

A

S

с

A

F

F

E

Proof.-CBD is a ▲ and CE and BA are two medians.
AFAB.

(85°, Cor.)

Bisecting FB gives the other point of division.

D

128°. Problem.-To construct a ▲ when the three medians

are given.

E

B

AD=1, BE=m, and CF=n,
AO=34, OB=3m, and OF=}n,

.. in the AOB we have two sides and the median to the

third side given. Thence AAOB is constructed by 126°

and 127°.

Constr.-Let AB be the segment.

Through A draw any line CD and make B AC AD. Bisect DB in E, and join CE, cutting AB in F.

Then AF is AB.

Then producing FO until OC=2FO, C is the third vertex of the triangle required.

B

E

Xx

m

n

Ex. To describe a square whose sides shall pass through

four given points.

Let P, Q, R, S be the given points, and suppose ABCD to be the square required. P Join P and Q upon opposite sides of the square, and draw QG || to BC. Draw SX I to PQ to meet BC in E, and draw EF || to CD. Then AQPG=AFSE,

and

SE=PQ.

с

Let l, m, n be the given medians, and suppose ABC to be the required triangle.

Then

OR

Hence the construction :-

Join any two points PQ, and through a third point S draw

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