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2. From centre B, with the same radius, cut the arc in C.
3. Join AC and BC. Then ABC is the required equilateral

triangle.
NOTE.—The three straight lines are all equal, since they are
radii of equal arcs.

Problem 19. To construct an equilateral triangle having a given height AB

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1. From the extremities of the line AB, draw CAD and

EBF perpendicular to it (Pr. 2). 2. From A as centre with any radius, describe a semicircle

cutting CAD in C and D. 3. From C and D with the same radius, cut the semicircle

in G and H. 4. From A draw lines through H and G, meeting EF in E

and F. Then A EF is the equilateral triangle required. NOTE. — The radius of a circle can be marked off six times round its circumference, hence the arc HG is 60°. Moreover, the three angles of a triangle, added together, are equal to two right angles, or 180° (Euc. I. 32). Hence, the angle HAB being 30°, and ABE 90°, the angle AEB is 60°.

Problem 20. To construct a triangle, the three sides AB, CD, and EF being given.

1. With centre A, and radius CD, describe arc GH.
2. With centre B, and radius EF, describe an arc cutting

GH in K.

3. Draw the straight lines AK, BK, then KAB is the

triangle required.

NOTE. "The greater side of every triangle is opposite to the greater angle” (Euc. I., 18). Hence angle AKB is greater than the angle KAB.

Problem 21.

To find the altitude of a given triangle ABC.

1. From point A, let fall the perpendicular AD (Pr. 3).
2. Line AE is the required altitude of the given triangle

ABC.

NOTE.-If the line A E does not fall on the base, the base must be produced, and then we can obtain the altitude of the triangle as above.

Problem 22.
To find the centre of a given triangle ABC.

1. Bisect any two of its angles, say, the angles at B and C

(Pr. 4). 2. Produce the bisecting lines, and let them meet in D.

Then D is the centre of the given triangle ABC.

NOTE. —Perpendiculars drawn from D to the three sides of the triangle are equal in length. They would thus become the radii of a circle which might be inscribed within the triangle (Euc. IV. 4.)

Problem 23. To construct a triangle, its base AB and the angles at the base A and B being given.

1. Draw line CD equal to AB.

2. Make angle C equal to angle A, and angle D equal to

angle B (Pr. 10). 3. Produce the sides until they meet in E. Then CED is

the required triangle.

Problem 24.

To construct a triangle, the altitude AB, and the two angles at the base, C and D, being given.

1. Through the point B, draw EF perpendicular to AB

(Pr. 2); also through A, draw GH perpendicular to

AB. 2. From point A draw AK, making the angle AKB equal

to angle C, by first making angle GAK equal to C (Pr. 10).

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3. In the same manner, make angle ALB equal to D. Then

AKL is the required triangle.

NOTE.— The angles GAK, AKB, are called alternate angles, and when a straight line falls upon two parallel straight lines, it makes the alternate angles equal to each other (Euc. I. 29).

Problem 25.

To construct a triangle, having its base AB, its altitude CD, and one side BC given.

1. Draw a line CE parallel to AB, at a distance from it

equal to the altitude CD (Pr. 8).

2. From

B, as centre, with radius BC, cut CE in the

point .

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8. Join CB, CA. Then ABC is the required triangle, and

CD drawn from C perpendicular to the base AB produced (Pr. 3) is the altitude.

Problem 26.

To construct a triangle on a given base AB, having angles of 60°, 30°, and 90°.

1. At B, construct a right angle, that is, raise a perpen

dicular (Pr. 2).

2. At A, make an angle of 60° (Pr. 7), and continue the

line, until it meets the perpendicular erected at B, in the point C. Then ABC is the required triangle.

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