To prove that P B is the bisector of the angle ABC: Draw PM and P N, perpendicular respectively to A B, and B C.

Then in the right triangles P B M and P B N, the side P B is common, and by hypothesis P M = PN.

Therefore the triangles are equal, and angle P B M = angle PBN. (Theorem XX).

Hence P B bisects angle A B C.

91. Scholium. The bisector of the angle is the locus of points which are within the angle and equally distant from its sides.

THEOREM XXVII.

92. If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, the third side of the first is greater than the third side of the second.

In the triangles A B C and D E F, let the side A B be equal to D E, and AC to D F, and let the included angle B A C be greater than angle E D F.

To prove that B C is greater than EF:

Place the triangle D E F in the position A B F', the side D E coinciding with its equal A B, and the vertex F falling at F'. Draw A G bisecting the angle C A F', and join F' G. Then in the triangles A C G and A F' G, the side A G is common, A C A F by hypothesis, and angle CAG = angle FAG by construction.

Therefore the triangles are equal in all respects (Theorem XVII), and F' G = C G.

But by axiom (a straight line being the shortest distance between two points), B G + F G is greater than B F'. Substituting for F G its equal C G,

BGCG is greater than BF'

That is, B C is greater than E F.

THEOREM XXVIII.

93. Conversely, if two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, the included angle of the first is greater than the included angle of the second.

In the triangles A B C and D E F, let the side A B be equal to D E, and A C to D F, and let B C be greater than E F. To prove that angle A is greater than angle D:

If angle A were equal to angle D, the triangles would be equal (Theorem XVII), and B C would be equal to E F.

And if angle A were less than angle D, B C would be less than E F (Theorem XXVII).

Both of these conclusions are contrary to the hypothesis;
Hence angle A is greater than angle D.

94. Scholium. The method of proof employed in Theorem XXVIII, is known as the "Reductis ad absurdum," or "indirect" method. The truth of the theorem is demonstrated by supposing it to be false, and showing that the hypothesis leads to a false conclusion.

QUADRILATERALS.

DEFINITIONS.

95. A Quadrilateral is a plane figure bounded by four straight lines, as A B C D.

96. A diagonal of a quadrilateral is a straight line joining two opposite vertices, as A C.

97. A Trapezium is a quadrilateral which has no two of its sides parallel; as A B C D.

98. A Trapezoid is a quadrilateral which has only two of its sides parallel; as E F G H.

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99. A Parallelogram is a quadrilateral whose opposite sides are parallel, as any rectangle or rhombus.

100. A Rectangle is a right-angled parallelogram, as IJ KL.

101. A Square is an equilateral rectangle, as M N O P.

102. A Rhomboid is an oblique-angled parallelogram, as Q R S T.

103. A Rhombus is an equilateral rhomboid, as U V W X.

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104. The bases of a trapezoid are its parallel sides; the altitude is the perpendicular distance between them.

105. The bases of a parallelogram are the side upon which it is supposed to stand and its parallel side; the altitude is the perpendicular distance between them.

THEOREM XXIX.

106. The opposite sides and angles of a parallelogram are equal to each other.

Let A B C D be a parallelogram ;

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To prove that A B D C, BC = A D, and the angles A and B respectively equal C and D :

Draw the diagonal B D. In the triangles A B D and B C D, the side B D is common. As B C and A D are parallel, the alternate interior angles C B D, and B D A are equal. (Theorem V.) And in the same way angle A B D = angle B D C.

Therefore the two triangles A B D and B D C are equal (Theorem XVIII;) and the sides opposite the equal angles are equal, i.e. A B D C, and B C A D; also the angle A = angle C and the angle

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A B C = angle A B D + DBC= BDC+BDA = ADC. 107. Corollary I. The diagonal divides a parallelogram into two equal triangles.

equal.

108. Corollary II.

Parallels included between parallels are

THEOREM XXX.

109. If two sides of a quadrilateral are equal and parallel the figure is a parallelogram.

Let A B C D be a quadrilateral having B C equal and parallel to A D;

To prove that A B C D is a parallelogram:

Draw the diagonal B D. As B C is parallel to A D, the alternate interior angles C B D and BDA are equal (Theorem V), therefore the two triangles C B D and B D A are equal (Theorem XVII); and the alternate interior angles A B D and BDC are equal. Therefore A B is parallel to DC (Theorem VI), and A B C D is a parallelogram.

110.

THEOREM XXXI.

The diagonals of a parallelogram bisect each other.

Let the diagonals of the parallelogram A B C D intersect at E.

To prove that A E

EC and BEED:

In the triangles A E D and B E C, we have A D = B C, (Theorem XXIX), also since the parallels A D and B C are cut by A C, angle E A D = angle E C B. And since the parallels A D and B C are cut by B D, angle E D A = angle E B C.

Therefore the triangles are equal (Theorem XVIII), and A E E C, and B E E D (homologous sides of equal triangles).

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