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PROPOSITION XV. THEOREM.

112. Two angles whose sides are parallel each to each, are either equal or supplementary.

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Let AB be parallel to EF, and BC to MN.

To prove ABC equal to

EHN, and to MHF, and

supplementary to ZEHM and to NHF.

Proof. Prolong (if necessary) BC and FE until they inter

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and

L DHN=LEDC.

(being ext.-int. 4 of || lines),

..LB=LDHN;

LB=LMHF (the vert. of DHN).

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Now DHN is the supplement of ▲ EHM and ▲ NHF.

..Z B, which is equal to Z DHN,

is the supplement of EHM and of ▲ NHF.

Q. E. D.

REMARK. The angles are equal when both pairs of parallel sides extend in the same direction, or in opposite directions, from their vertices; the angles are supplementary when two of the parallel sides extend in the same direction, and the other two in opposite directions, from their vertices.

PROPOSITION XVI. THEOREM.

113. Two angles whose sides are perpendicular each to each, are either equal or supplementary.

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Let AB be perpendicular to FD, and AC to GI.

To prove

ZDFI.

BAC equal to Z DFG, and supplementary to

Proof. Suppose AK drawn to AB, and AHL to AC.

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(two angles are equal whose sides are || and extend in the same direction

from their vertices).

The Z BAK is a right angle by construction.
..ZBAH is the complement of Z KAH
The CAH is a right angle by construction.
..ZBAH is the complement of Z BAC.

:. L BAC=Z KAH,
(complements of equal angles are equal).

.. LDFGL BAC.

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DFG, is also the supplement

:. ZDFI, the supplement of

of

BAC.

Q. E. D.

REMARK. The angles are equal if both are acute or both obtuse; they

are supplementary if one is acute and the other obtuse.

PERPENDICULAR AND OBLIQUE LINES.

PROPOSITION XVII. THEOREM.

114. The perpendicular is the shortest line that can be drawn from a point to a straight line.

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Let AB be the given straight line, P the given point, PC the perpendicular, and PD any other line drawn from P to AB.

To prove PC PD.

Proof. Produce PC to P', making CP' PC; and draw DP'. On AB as an axis, fold over CPD until it comes into the plane of CP'D.

But

The line CP will take the direction of CP',
(since ▲ PCD = ▲ P'CD, each being a rt. ▲ by hyp.).
The point P will fall upon the point P',

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.. PD+P'D=2 PD,

and PC + CP' = 2 PC.

PC+CP' <PD+DP',

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

Cons.

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115. SCHOLIUM. The distance of a point from a line is understood to mean the length of the perpendicular from the point to the line.

PROPOSITION XVIII. THEOREM.

116. Two oblique lines drawn from a point in a perpendicular to a given line, cutting off equal distances from the foot of the perpendicular, are equal.

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

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

FA will take the direction of FO,

(since ▲ CFA = ≤ CFO, each being a rt. ▲ by hyp.).

Point A will fall upon point O,

(since FA FO by hyp.).

=

.. line CA = line CO,

(their extremities being the same points).

Q. E. D.

117. COR. Two oblique lines drawn from a point in a perpendicular to a given line, cutting off equal distances from the foot of the perpendicular, make equal angles with the given line, and also with the perpendicular.

PROPOSITION XIX. THEOREM.

118. 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 CB be two lines drawn from the point C to the extremities of the straight line AB. Let OA and OB be two lines similarly drawn, but included by CA and CB.

Το prove

Proof.

Then

and

CA+ CB > OA + OB.

Produce AO to meet the line CB at E.

AC+ CE> OA+ OE,

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

BE+OE> BO.

Add these inequalities, and we have

CA+CE+ BE+OE> OA+OE+OB.

Substitute for CE+ BE its equal CB,

and take away OE from each side of the inequality.

We have

CA+CB > OA+OB.

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Q. E. D.

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