Sol.—If X be a right angle, describe a semicircle on the given line, and the thing required is done; for the angle in a semicircle is a right angle.
If not, make with the given line AB the angle BAE equal to X. Erect AC at right angles to AE, and BC at right angles to AB. On AC as diameter describe a circle: it will be the circle required.
Dem.—The circle on AC as diameter passes through B, since the angle ABC is right [xxxi.] and touches AE, since the angle CAE is right [xvi.]. Therefore the angle BAE [xxxii.] is equal to the angle in the alternate segment; but the angle BAE is equal to the angle X (const.). Therefore the angle X is equal to the angle in the segment described on AB.
Exercises.
1. Construct a triangle, being given base, vertical angle, and any of the following data:—1. Perpendicular. 2. The sum or difference of the sides. 3. Sum or difference of the squares of the sides. 4. Side of the inscribed square on the base. 5. The median that bisects the base.
2. If lines be drawn from a fixed point to all the points of the circumference of a given circle, the locus of all their points of bisection is a circle.
3. Given the base and vertical angle of a triangle, find the locus of the middle point of the line joining the vertices of equilateral triangles described on the sides.
4. In the same case, find the loci of the angular points of a square described on one of the sides.
PROP. XXXIV.—Problem.
To cut off from a given circle (ABC) a segment which shall contain an angle equal to a given angle (X).