shown. Join FA, EB, and DC. If these lines are produced, they should pass through the points H, L, and K, respectively. Complete the figure by drawing the circles as shown.

5. Draw two lines OA, OA, each 10 inches long, and containing an angle AOA of 88°. Measure accurately and state the distance apart of A, A. What should be the exact distance AA ? State any special precautions you may have taken to ensure accuracy in your construction. [B.E.]

6. Take from the tables the chord and tangent of 22°. Construct an angle of 22° by using the value of the chord, and a second angle of 220 by using the tangent. From each of these angles determine the sine of 22°, and compare the mean of the two values with the true value of sine 22°. [B.E.]

7. By aid of your protractor, and without using the tables, find the value of

sin 23°+tan 23° + cos 23°.

H

FIG. 10.

Now take out the true values of the sine, tangent, and cosine from the tables, add them, and calculate the percentage error in your first answer. [B.E.]

8. In working this question employ a decimal scale of inch to 1 unit. Draw a circular arc, radius 10 units, centre O. Mark a chord AB of this arc 3:47 units long, and draw the radii OA, OB. Measure the angle AOB in degrees.

From B draw a perpendicular BM on OA, and at A draw a tangent to meet OB produced in N. Measure carefully BM, AN and the arc AB (on the above unit scale) and calculate the sine and tangent of the angle, and the angle in radians.

Give the correct answers for the degrees, sine, tangent, and radians, the numbers being taken directly from

to be collinear. Points which lie on the circumference of a circle are said to be concyclic. A straight line drawn across a system of lines is called a transversal.

The three straight lines which join the angular points of a triangle and the middle points of the opposite sides are called the medians of

the triangle, and the point at which the medians intersect is called the centroid of the triangle.

The three perpendiculars drawn from the angular points of a triangle to the opposite sides meet at a point called the orthocentre of the triangle, and the triangle formed by joining the feet of the perpendiculars is called the pedal triangle, or orthocentric triangle of the original triangle.

If a straight line be be drawn to cut the sides AC and AB of a triangle ABC (Fig. 12) at b and c respectively, so that the angle Acb is equal to the angle ACB, then be with reference to AB and AC is said to be antiparallel to BC. An antiparallel be is parallel to the tangent AD to the circumscribing circle at A. The middle points of the antiparallels to one side of a triangle lie on a straight line which passes through the opposite angular point of the triangle, and this line is called a symmedian of the triangle. The three symmedians of a triangle intersect at a point called the symmedian point of the triangle. 7. Equality of Two Triangles.--A triangle has six parts, three sides and three angles, and, except in two cases, a triangle is completely determined when three of its six parts are known. Another way of stating the above is that if there be two triangles having three parts in the one equal respectively to the corresponding parts in the other, then, except in two cases, the two triangles are equal in every respect. The different cases are as follows:

:

Two triangles are equal in every respect when

FIG. 12.

(1) The three sides of the one are equal to the three sides of the other, each to each.

(2) Two sides and the included angle in one are equal to two sides and the included angle in the other, each to each.

(3) Two angles and the side adjacent to them in one are equal to two angles and the side adjacent to them in the other, each to each.

(4) Two angles and the side opposite to one of them in one are equal to two angles and the side opposite to one of them in the other, each to each, the equal sides being opposite to equal angles.

These four cases are illustrated in Fig. 13; the corresponding equal parts in the two triangles have similar marks on them.

Of the two exceptional cases mentioned above, one is that in which the three angles of one triangle are equal to the three angles of the other. In this case, the two triangles have the same shape, but not necessarily the same size. The other case is that in which two sides and an angle opposite to one of them in one triangle are equal to

two sides and an angle opposite to one of them in the other, cach to each, the equal angles being opposite to equal sides. In this case the triangles may be equal, but they are not necessarily so, as is shown in Fig. 14, where ABC, is one triangle and ABC, the other.

FIG. 14.

B

8. To bisect the angle between two straight lines. First, let the two straight lines OA and OB intersect at a point O which is accessible (Fig. 15). With O as centre and any radius describe arcs to cut OA at C and OB at D. With centres C and D describe intersecting arcs of equal radii. The straight line OE, which joins O with the point of intersection of these arcs, will bisect the angle AOB.

An important property of the bisector of an angle is that the perpendiculars, such as FH and FK, drawn from any point on it to the lines containing the angle are of equal length.

Next, let the given lines PQ and RS (Fig. 16) meet at a point which is inaccessible. Draw any straight line MN to cut PQ at M and RS at N. Bisect the angles PMN and RNM by straight lines which intersect at L. Through M and N draw perpendiculars to LM and LN respectively, and let them meet at T. The straight line LT will if produced meet PQ and RS produced at the same point, and will bisect the angle between PQ and RS. It is easy to see that MT and NT bisect the angles NMQ and MNS respectively, which suggests another way of drawing these lines.

9. To find a point on a given straight line such that the sum of its distances from two given points shall be the least possible. Let LN (Fig. 17) be the given straight line and A and B the two given points. The given points are on the same side of LN. Draw ACD at right angles to LN, cutting LN at C. Make CD equal to AC. Join DB, cutting LN at E. E is the point required.

Join AE.

DK

FIG. 17.

B

N

Comparing the triangles ACE and DCE, it is obvious that AE is equal to DE. Hence AE + BE = DB. If any other point F in LN

be taken and F be joined to A, B, and D, it is evident that AF is equal to DF, and AF + BF = DF + BF. But DF+ BF is greater than DB, therefore AF + BF is greater than AE + BE.

It is evident that the angles AEL and BEN are equal, and the problem may be stated in another way, namely, to find a point E in a given line LN such that the lines joining E to two given points A and B shall be equally inclined to LN.

If LN represents a mirror, and a ray of light from A is reflected from the mirror and passes through B, then the incident ray will strike the mirror at the point E found by the above construction.

If the points A and B are on opposite sides of LN it is obvious that E will be the point where the straight line AB cuts LN. In this case, however, there is no corresponding case of reflection from a mirror.

Referring to Fig. 17, and considering LN to be a mirror, D is the image of the point A in the mirror.

да

B

α

C

10. Similar Rectilineal Figures. -Two rectilineal figures are said to be similar when they have their several angles equal, each to each, and the sides about their equal angles proportionals. For example, the figure ABCD (Fig. 18) is similar to the figure abcd A when the angles A, B, C, and D of the one are equal to the angles a, b, c, and d, respectively, of the other, and the ratios da ab be cd equal to the ratios and respectively. Stated in another al' be' cd' da way, the similar figures have the same shape but have different dimensions.

FIG. 18.

DA AB BC
AB' BC' CD'

CD

and

are

DA

When the two figures are triangles, they are similar if the angles The of the one are equal to the angles of the other, each to each. equality of the ratios of the sides about the equal angles follows as a consequence of the equality of the angles.

It is important to remember that the ratio of the areas of two similar figures is equal to the ratio of the squares on their correspond. ing sides. For example, referring to Fig. 18, area of ABCD: area of abcd :: AB2: ab2.

11. Through a given point to draw a straight line which shall be concurrent with two given straight lines when the point of concurrence is inaccessible.AB and CD (Fig. 19) are the given lines, and P is the given point. Draw a triangle PEF having one angular point at P and the other angular points one on AB and one on CD.

K

FIG. 19.

Draw H parallel to EF, meeting AB at H and CD at K. Through Hand draw parallels to EP and FP respectively, and let them

meet at Q. A straight line through P and Q will, if produced, meet AB and CD produced at the same point.

12. Graphic Arithmetic.-The solution of ordinary arithmetical problems by geometrical constructions is of some interest when considered as a section of practical geometry, but graphic arithmetic, when regarded as a means to an end, is in general a poor substitute for the ordinary method by calculation. The subject of graphic arithmetic will therefore be treated somewhat briefly here.

Representation of numbers by lines. If the line AB, Fig. 20 (a), be taken to represent the number one, then a line CD n times the length of AB will represent the number n. Again, if AB represents the number one, the number represented by the line EF will be the number of times that EF contains AB. In the above examples AB is called the unit.

Ordinary drawing scales may be used in marking off or measuring lengths which represent numbers. Scales which are decimally divided are the most convenient for this work.

Addition and subtraction. To add a series of numbers together. Draw a straight line OX, Fig. 20 (b), of indefinite length. Fix upon a unit, that is, decide what length shall represent the number one. Mark a definite point O on OX. Make OA = the first number, AB the second number, BC: the third number, and so on. From O to the last point determined in this way will be the answer required.

=

=

From one number to subtract another. Make OA, Fig. 20 (c), measured to the right of O, equal to the first number. Make AB, measured to the left of A, equal to the second number which is to be subtracted from the first. From O to B is the answer required. If OB is to the right of O the answer is positive (+), and if OB is to the left of O, the answer is negative (−).

· e,

To find the value of such an expression as a + b -c+d where a, b, c, d, and e represent numbers, make (Fig. 20 (d)) OA = a, AB = b, BC = c, CD = d, and DE =e, then OE = a + b -c+d-e. Note that in adding the lengths are measured to the right, while in subtracting they are measured to the left.

Proportion. In Figs. 21 and 22, OX and OY are two straight lines making any convenient angle with one another. AB and CD are parallel lines meeting OX at A and C, and OY at B and D. The triangles AOB and COD are similar, and therefore OA: OB:: DC: OD,