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PROBLEM II.

To bisect a given angle A B C. From the angular point B, with any radius, describe the arc A C, and from the points A and C, with the same, or any B other radius, describe arcs cutting each other in n. Join B n, and it will bisect the angle A B C. For join A n, C n, then as A B is equal to B C, A n to

А,
C n, and B n is common to both the triangles A B n and
C Bn; therefore the angle A B n is equal to the angle
CBn; or A B C is bisected by B n.

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PROBLEM III.

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From a given point in a given line A B to draw a perpendicular. 1st. When the point is near the middle of the line.

On each side of the point C take any two equal distances Cn, Cm; and from n and m with any radius greater than C n or C m, describe arcs cutting each other in s. Through s and C draw the line s C, and it will be the required perpendicular. For join s m, s n; then as these two lines are

C equal, n C equal to m C, and Cs common to the triangles n C s, m Cs, the angles m Cs and n C s are equal ; and therefore s C is perpendicular to A B.

2d. When the point is at or near the end of the line.

Take any point o as a centre, and with the radius o C describe an arc m C n, cutting A B in m and C. Through m and o draw the line m on cutting the arc m Cn in n. Join n C, and it will be the required perpendicular.

А. For the angle m C n being an angle in a semicircle, is a right angle, and therefore n C is perpendicular to A B.

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PROBLEM IV.

From a given point C out of a given line A B, to draw a line perpendicular to A m.

1st. When the given point is nearly opposite the middle of the line.

Take any point o on the other side of the line A B, and from the centre C with the radius C o describe an arc no m cutting A B in n and m. From n, m, with the same, or any other radius, describe arcs cutting each other in s. Through the points C, s,

-B draw the line C Gs, and C G will be the perpendicular required.

For let n C, ns, m C, m s, be joined; then it

may be shewn as in the demonstration of Prob. I. that the angles A G C and B G C are equal, and C G is consequently perpendicular to A B.

2d. When the point is nearly opposite the end of the line.

To any point m in the line A B draw the line C m, and bisect it in the point n. From n, with the radius n m or n C

C describe the arc C Gm cutting A B in G. Through the point C draw the line C G and it will be the perpendicular required. For A G b being an 'angle in a semicircle is a

A

-B right angle, and therefore C G is perpendicular to A B.

PROBLEM V.

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At a given point D in a given line D E to make an angle equal to a given angle ABC.

From B as a centre with any radius describe the arc n m cutting B A, B C in the points m, n; and

C

F from D with the same radius describe the arcs r s.

Take the distance m n and apply it to the arc r s from r to

B

mm A A s. Through the points D s draw the line D F, then the angle E D F will be equal to the angle A B C.

For let m, n and s, r be joined; then the triangles B n m and Dsr having B n and Bm equal to Ds and Dr, and n m equal to s r, therefore the angles B and D are equal.

PROBLEM VI.

Through a given point C, to draw a line parallel to a given line A B.

From C to A B draw any line C D; then through C draw the line C E, making the angle D C E equal to the angle

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E C DA; and C E will be the required line.

For the alternate angles A D C and DC E being equal, the lines A D and C E are parallel.

A DB

PROBLEM VII.

To draw a line parallel to a line A B, given in position, at a distance from it equal to another given line C D.

At any point E in A B, erect the perpendicular EF, and make it equal to C D. Through the point F draw F G parallel F G to A B; then F G will be the required line. For F G is parallel to A B by construction ; and

A E FE, its distance from A B, is equal to the given

PROBLEM VIII.

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e

d

A:

To divide a given line A B into any proposed number of equal parts.

From A, one end of the given line, draw the line A m making any angle with A B; and from B, the other end, draw B r parallel to Am. In each of these lines, beginning at A and B, set off as many equal parts A a, a b, bc, &c. and Bf, fg,gh,

B &c. as A B is to be divided into. Join the points B e, df, cg, &c. and these lines will divided A B as required.

For as d e is equal and parallel to Bf, B e and df are parallel ; and in the same manner may cg, b h, and a i be shewn to be parallel to Be. Hence the parts into which A B is divided are proportional to the corresponding parts of A e; and as the parts of A e are equal to each other, the parts of A B are equal to each other also.

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PROBLEM IX.

To find the centre of a given circle A B C. Draw any chord A B, and bisect it with the perpendicular C D. Bisect C D in 0, and O will be the centre required. For if the centre is in the line C D it must be in

F 0, that point in the chord which is equally distant

01 from its extremities. If however the centre is not in C D, let it be in F any point out of that line ; and

A

E B join F E, F A, and F B. Then will A F be equal to FB, A E to E B, and E F common to the triangles A E F and BE F; the angles A E F and B E F will therefore be equal, and consequently each of them will be a right angle. Hence the angles CEF and FE B will be equal, the less to the greater, which is impossible. The centre of the circle is therefore in the line C D, and it is consequently in the point o.

PROBLEM X.

To describe the circumference of a circle through three given points A, B, C.

Join A C and B C, and bisect these lines with the perpendiculars DO and E O, and from the point O with the dis

С tance O A, O B, or 0 C, describe the circle ABC, and it will be the circle required.

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АВ For as AD and DO are respectively equal to C D and DO, and the included angles A D C and CD O are equal, A O is equal to CO; and for a

B

like reason B O is equal to C O. Hence O is the centre of a circle passing through A, B, and C.

Remark. By this problem a circle may be described about a triangle.

PROBLEM XI.

To inscribe a circle in a given triangle A B C. Bisect any two of the angles, as A and B, and the bisecting lines, AD and D B, will meet in D, the centre of the circle.

F For from D draw DE, DF, and DG, perpendi

DI culars to A B, BC, and A C. Then, as the angles

A E B DAG and DAE are equal by construction, and the angles ADG and ADE are equal ; and A D adjacent to the equal angles in each triangle; the triangles are identical, and therefore D E and D G are equal ; and thus also may D F be shewn to be equal to DE or DG; and consequently, as the angles at E, F, and G, are right angles, a circle described from D, with the distance DE, DF, or D G, will touch A B, BC, and A C, the sides of the triangle; and the circle will therefore be inscribed in the triangle.

PROBLEM XII.

On a given line A B to describe the segment of a circle, which will contain an angle equal to a given angle, as M.

From A draw A C, making the angle CAB equal to the given angle M. Bisect A B with

E

E the perpendicular DE; and draw

АЎ D B A D B В A E perpendicular to AC, meeting DE in E; then from the centre E, with the radius A E or

M

M E B, describe a circle, and the segment of it, alternate to the angle C AB, will be the segment required.

For as A D is equal to D B, and the angles ADE and B D E are equal, and D E common to the triangles A D E and B D E; therefore A E and B E are equal. And as A E is perpendicular to AC, therefore A C is a tangent to the circle of which A B is a chord; and hence the angle C A B is equal to any angle in the alternate segment.

Remark. If the given angle is a right angle, the required segment will be a semicircle described on the given line.

PROBLEM XIII.

To make a plain diagonal, decimal scale.

C

E 3

А

B

2

scale, and draw ten lines paral

F lel to it, and equidistant from each other; draw A D and BC perpendicular to A B; and divide A B and C D each into ten equal parts. Join B to the first division in CD, the first division on AB to the second on CD, &c. In A B produced take B1, 12, &c. each equal to AB; and from the points 1, 2, &c. draw lines parallel to B C, meeting CD produced; then the scale will be completed.

For A B is divided into ten equal parts, and the oblique lines drawn from A B to C D subdivide the lines drawn parallel to A B again into tenths of the divisions of A B. For as B C is equal to ten times the distance of each of the parallels, therefore the first division on CD is ten times the part of the first parallel from A B, intercepted by BC, and the oblique line drawn from B; the part intercepted by the same lines on the second parallel is double the part intercepted on the first, &c.; or the intersection of the diagonal lines with the parallels exhibit the decimal divisions of the parts of A B.

EXAMPLES FOR EXERCISE IN GEOMETRICAL

INVESTIGATION.

1. From two given points, to draw two equal straight lines, which shall meet in the same point, in a line given in position.

2. From two given points on the same side of a line, given in position to draw two lines which shall meet in that line, and make equal angles with it.

3. If, from a point without a circle, two straight lines be drawn to the concave part of the circumference, making equal angles with the line joining the same point and the centre, the parts of these lines which are intercepted within the circle are equal.

4. Of all straight lines which can be drawn from two given points to meet on the convex circumference of a given circle, the sum of those two will be the least which make equal angles with the tangent at the point of concourse.

5. 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.

6. From two given points on the same side of a line given in position, to draw two straight lines which shall contain a given angle, and be terminated in that line.

7. If on each side of any point in a circle any number of equal arcs be taken, and the extremities of each pair be joined, the sum of the chords so drawn, will be equal to the last chord produced to meet a line drawn from the given point through the extremity of the first arc.

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