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119. Of two oblique lines drawn from the same point in a perpendicular, cutting off unequal distances from the foot of the perpendicular, the more remote is the greater.

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Let OC be perpendicular to AB, OG and OE two oblique lines to AB, and CE greater than CG.

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(two oblique lines drawn from a point in a 1, cutting off equal distances from the foot of the 1, are equal).

Prolong OC to D, making CD OC.

Draw ED and FD.

Since AB is to OD at its middle point,

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(the sum of two oblique lines drawn from a point to the extremities of a straight line is greater than the sum of two other lines similarly drawn, but included by them).

.. 20E> 20F, or OE > OF.

But OFOG. Hence OE> OG.

Q. E. D.

120. COR. Only two equal straight lines can be drawn from a point to a straight line; and of two unequal lines, the greater cuts off the greater distance from the foot of the perpendicular.

PROPOSITION XXI. THEOREM.

121. Two equal oblique lines, drawn from the same point in a perpendicular, cut off equal distances from the foot of the perpendicular.

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Let CF be the perpendicular, and CE and CK be two equal oblique lines drawn from the point C to AB.

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Proof. Fold over CFA on CF as an axis, until it comes into the plane of CFB.

The line FE will take the direction FK,

CFE CFK, each being a rt. Z by hyp.).

=

Then the point E must fall upon the point K,

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Otherwise one of these oblique lines must be more remote from the perpendicular, and therefore greater than the other; which is contrary to the hypothesis that they are equal. § 119

Q. E. D.

Ex. 11. Show that the bisectors of two supplementary adjacent angles are perpendicular to each other.

Ex. 12. Show that the bisectors of two vertical angles form one straight line.

Ex. 13. Find the complement, and the supplement, of angle containing 26° 52′ 37′′.

PROPOSITION XXII. THEOREM.

122. Every point in the perpendicular, erected at the middle of a given straight line, is equidistant from the extremities of the line, and every point not in the perpendicular is unequally distant from the extremities of the line.

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Let PR be a perpendicular erected at the middle of the straight line AB, O any point in PR, and C any point without PR.

Draw OA and OB, CA and CB.

To prove OA and OB equal, CA and CB unequal.

Proof.

PA= PB.

..OA= OB,

Hyp.

$ 116

(two oblique lines drawn from the same point in a 1, cutting off equal distances from the foot of the 1, are equal).

Since C is without the perpendicular, one of the lines, CA or CB, will cut the perpendicular.

Let CA cut the at D, and draw DB.

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(two oblique lines drawn from the same point in a 1, cutting off equal distances from the foot of the 1, are equal).

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(a straight line is the shortest distance between two points). Substitute in this inequality DA for DB, and we have

That is,

CB CD+DA.

CB < CA,

Q. E. D.

123. Since two points determine the position of a straight line, two points equidistant from the extremities of a line determine the perpendicular at the middle of that line.

THE LOCUS OF A POINT,

124. If it is required to find a point which shall fulfil a single geometric condition, the point will have an unlimited. number of positions, but will be confined to a particular line, or group of lines.

Thus, if it is required to find a point equidistant from the extremities of a given straight line, it is obvious from the last proposition that any point in the perpendicular to the given line at its middle point fulfils the condition, and that no other point does; that is, the required point is confined to this perpendicular. Again, if it is required to find a point at a given distance from a fixed straight line of indefinite length, it is evident that the point must lie in one of two straight lines, so drawn as to be everywhere at the given distance from the fixed line, one on one side of the fixed line, and the other on the other side.

The locus of a point under a given condition is the line, or group of lines, which contains all the points that fulfil the given condition, and no other points.

125. SCHOLIUM. In order to prove completely that a certain line is the locus of a point under a given condition, it is necessary to prove that every point in the line satisfies the given condition; and secondly, that every point which satisfies the given condition lies in the line (the converse proposition), or that every point not in the line does not satisfy the given condition (the opposite proposition).

126. COR. The locus of a point equidistant from the extremities of a straight line is the perpendicular bisector of that line.

§§ 122, 123

TRIANGLES.

127. A triangle is a portion of a plane bounded by three straight lines; as, ABC.

The bounding lines are called the sides of the triangle, and their sum is called its perimeter; the angles formed by the sides are called the angles of the triangle, and the vertices of these angles, the vertices of the triangle.

B

D

FIG. 1.

128. An exterior angle of a triangle is an angle formed between a side and the prolongation of another side; as, ACD. The interior angle ACB is adjacent to the exterior angle; the other two interior angles, A and B, are called oppositeinterior angles.

B

Scalene.

Isosceles.

Equilateral.

129. A triangle is called, with reference to its sides, a scalene triangle when no two of its sides are equal; an isosceles triangle, when two of its sides are equal; an equilateral triangle, when its three sides are equal.

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130. A triangle is called, with reference to its angles, a right triangle, when one of its angles is a right angle; an obtuse

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