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PROPOSITION IV. THEOREM.

53. Two oblique lines drawn from a point in a perpendicular, cutting off equal distances from the foot of the perpendicular, are equal.

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Let FC be the perpendicular, and CA and C O two oblique lines cutting off equal distances from F.

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Fold over CFA, on C F as an axis, until it comes into the

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PROPOSITION V. THEOREM.

54. The sum of two 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.

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Let CA and C B be two lines drawn from the point C to the extremities of the straight line A B. Let O A and O B be two lines similarly drawn, but included by CA and C B.

We are to prove

CA+C BOA + O B.

Produce AO to meet the line CB at E.

Then

ACCE> AO+ OE,

(a straight line is the shortest distance between two points),

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Add these inequalities, and we have

CA+CE + BE+OE> OA+OE + O B.

Substitute for CE + BE its equal C B,

and take away OE from each side of the inequality.

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Q. E. D.

PROPOSITION VI. THEOREM.

55. 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 CF be perpendicular to A B, and C K and C H two oblique lines cutting off unequal distances from F.

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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).

But

CH+HECK + KE,

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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);

..2 CH2CK;

..CH>CK.

Q. E. D.

56. COROLLARY. 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 VII. THEOREM.

57. 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 C F be the perpendicular, and C E and C K be two equal oblique lines drawn from the point C.

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

The line FE will take the direction FK,

=

(CFE LCF K, each being a rt. 4).

Then the point E must fall upon the point K;

otherwise one of these oblique lines must be more remote from the 1,

and .. greater than the other; which is contrary to the hypothesis.

.. FE=FK.

§ 55

Q. E. D.

PROPOSITION VIII. THEOREM.

58. If at the middle point of a straight line a perpendicular be erected,

I. Any point in the perpendicular is at equal distances from the extremities of the straight line.

II. Any point without the perpendicular is at unequal distances from the extremities of the straight line.

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

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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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We are to prove CA and C B unequal.

One of these lines, as CA, will intersect the 1.

From D, the point of intersection, draw D B.

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