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2. From centre B, with the same radius, cut the arc in C.
Problem 19. To construct an equilateral triangle having a given height AB
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.
GH in K.
3. Draw the straight lines AK, BK, then KAB is the
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.
To find the altitude of a given triangle ABC.
1. From point A, let fall the perpendicular AD (Pr. 3).
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.
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.
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).
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).
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).
B, as centre, with radius BC, cut CE in the
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.
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.