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78°. Theorem.—If a line cuts a given line it cuts every parallel to the given line.

P

M

L

N

Let L cut M, and let N be any parallel to M. Then L cuts N.

Proof.-If L does not cut N it is || to N..

But M is to N. Therefore through the same point P two lines L and M pass which are both || to N.

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79°. Theorem.—If a transversal to two lines makes the sum of a pair of interadjacent angles less than a straight angle, the two lines meet upon that side

of the transversal upon which these interadjacent angles lie.

G

A

L

E

K

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H

GH is a transversal to AB and CD,

LBEF+LEFD < L.

Then AB and CD meet towards B and D.

Proof. Let LK pass through E making 4KEF=LEFC.

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(78°)

Again, ·.· EB lies between the parallels, and AE does not, the point where AB meets CD must be on the side BD of the transversal.

q e.d.

Cor. Two lines, which are respectively perpendicular to two intersecting lines, intersect at some finite point.

\A

D

B

80°. Def.-1. A closed figure having four lines as sides is in general called a quadrangle or quadrilateral.

Thus ABCD is a quadrangle.

2. The line-segments AC and BD which join opposite vertices are the diagonals of the quadrangle.

3. The quadrangle formed when two parallel lines intersect two other parallel lines is a parallelogram, and is usually denoted by the symbol

81°. Theorem.—In any parallelogram—

1. The opposite sides are equal to one another.

2. The opposite internal angles are equal to one another.

3. The diagonals bisect one another.

AB is to CD, and AC is || to BD, and

AD and BC are diagonals.

1. Then AB=CD and AC=BD.

A

D

B

Proof.. AD is a transversal to the parallels AB and CD,

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and ·.· AD is a transversal to the parallels AC and BD,

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(74°, 1)

(74°, 1)

(59°)

q.e.d.

2. LCAB=4BDC and LACD=LDBA.

Proof. It is shown in I that LCAD=LADB and BAD

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82°. Def. 1.—A parallelogram which has two adjacent sides

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Cor. 2. Since AC is the right bisector of BD,

and BD the right bisector of AC,

C

D

(54°)

Therefore the diagonals of a rhombus bisect one another

at right angles.

Def. 2.—A parallelogram which has one right angle is a rectangle, and is denoted by the symbol .

Cor. 3. Since the opposite angle is a

and the adjacent angle is a,

Therefore a rectangle has all its angles right angles.

(81°, 2)

(74°, 3)

Cor. 4. The diagonals of a rectangle are equal to one another.

Def. 3.-A rectangle with two adjacent sides equal is a square, denoted by the symbol ☐.

Cor. 5. Since the square is a particular form of the rhombus and a particular form of the rectangle,

Therefore all the sides of a square are equal to one another; all the angles of a square are right angles; and the diagonals of a square are equal, and bisect each other at right angles.

84°. Theorem.--If three parallel lines intercept equal segments upon any one transversal they do so upon every transversal.

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AE is a transversal to the three parallels AB, CD, and EF, so that AC=CE, and BF is any other transversal. Then BI) =DF.

Proof. Let GDH passing through D be to AE.

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Def.-The figure ABFE is a trapezoid.

Therefore a trapezoid is a quadrangle having only two

sides parallel. The parallel sides are the major and minor bases of the figure.

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Or, the line-segment joining the middle points of the nonparallel sides of a trapezoid is equal to one-half the sum of the parallel sides.

Cor. 2. When the transversals meet upon one of the extreme parallels, the figure AEF' becomes a ▲ and CD' becomes a line passing through the middle points of the sides AE and AF', and parallel to the base EF'.

E

C

D'

Therefore, 1, the line through the middle point of one side of a triangle, parallel to a second side, bisects the third side. And, 2, the line through the middle points of two sides of a triangle is parallel to the third side.

85°. Theorem.-The three medians of a triangle pass through a common point.

CF and AD are medians intersecting in O.
Then BO is the median to AC.

Proof. Let BO cut AC in E, and let

AG || to FC meet BO in G. Join CG. Then, BAG is a ▲ and FO passes through the middle of AB and is || to AG,

F

B

(84°, Cor 2)

A

E

C

G

.. O is the middle of BG.
Again, DO passes through the middle

points of two sides of the CBG,

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Def.-When three or more lines meet in a point they are said to be concurrent.

Therefore the three medians of a triangle are concurrent.

Def. 2.—The point of concurrence, O, of the medians of a triangle is the centroid of the triangle.

Cor. Since O is the middle point of BG, and E is the middle point of OG, (81°, 3)

OE=OB,
=EB.

Therefore the centroid of a triangle divides each median at two-thirds of its length from its vertex.

A

86°. Theorem.-The three right bisectors of the sides of a

N

B

triangle are concurrent.

Proof.-Let L and N be the right bisectors of BC and AB respectively.

Then L and N meet in some point O.

(79°, Cor.) Since L is the right bisector of BC, and N of AB, O is equidistant from B and C, and is also equidistant from A and B.

(53°)

Therefore O is equidistant from A and C, and is on the right bisector of AC.

Therefore the three right bisectors meet at O.

(54°)

q.e.d.

Cor. Since two lines L and N can meet in only one point (24°, Cor. 3), O is the only point in the plane equidistant from A, B, and C.

Therefore only one finite point exists in the plane equidistant from three given points in the plane.

Def. The point O, for reasons given hereafter, is called the circumcentre of the triangle ABC.

87°. Def.-The line through a vertex of a triangle perpendicular to the opposite side is the perpendicular to that side, and the part of that line intercepted within the triangle is the altitude to that side.

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