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CHAPTER V.

INEQUALITIES,

141. The symbol > is called the Sign of Inequality, and a > b means that a is greater than b; so a < b means that a is less than b.

THEOREM X.

142. An exterior angle of a triangle is greater than either of the two opposite interior angles.

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HYPOTHESIS. Let the side AB of the triangle ABC be produced to D. CONCLUSIONS. X CBD> * BCA.

* CBD > X BAC. PROOF. By 136, bisect BC in H.

By 100 and 101, join AH, and produce it to F; making, by 133, HF = AH.

By 100, join BF.

CH =

Then, in the triangles AHC and FHB, by construction,
HB, AH =

HF, and

* AHC * FHB;
(108. Vertical angles are equal.)
..

A AHC = A FHB, (124. Triangles are congruent if two sides and the included angle are equal in each.)

i. X HCA = * HBF. Now, % CBD > * HBF,

(85. The whole is greater than its part.)

.. CBD>X HCA. Similarly, if CB be produced to G, it may be shown that

* ABG>X BAC, is the vertical % CBD > * BAC.

THEOREM XI.

143. Any two angles of a triangle are together less than a straight angle.

B

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HYPOTHESIS. Let ABC be any A.
CONCLUSION. X A+ * B < st. 4.
PROOF. Produce CA to D.

Then * BAD> <B, (142. An exterior angle of a triangle is greater than either opposite interior angle.)

: BAD + % BAC > X B + BAC,
:. st. 4.>4B + X A.

144. COROLLARY I. No triangle can have more than one right angle or obtuse angle.

145. COROLLARY II. There can be only one perpendicular from a point to a line.

THEOREM XII.

146. If one side of a triangle be greater than a second, the angle opposite the first must be greater than the angle opposite the second.

B

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HYPOTHESIS. A ABC, with side a > side c.
CONCLUSION. BAC>* C.
PROOF. By 133, from a cut off BD = c.
By 100, join AD.
Then, because BD = 6, .. * BDA = * BAD.
(126. In an isosceles triangle the angles opposite the equal sides are equal.)

And, by 142, the exterior X BDA of A CDA > the opposite interior X C,

also X BAD> *C. Still more is X BAC > * C.

147. COROLLARY. If one side of a triangle be less than the second, the angle opposite the first will be less than the angle opposite the second. For, if a <b, .:b>a; :. by 146, X B>XA, :. * A<x D. 148. INVERSES. We have now proved, By 146, if a > b,

x 4 > x B ; By 126, if a = b, .. A= % B; By 147, if a < b,

* A< *B. Therefore, by 33, the inverses are necessarily true, namely, If X A > XB,

a > b; If X A = * B, :.

If % A < * B, 149. COROLLARY. An equiangular triangle is also equilateral.

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a < b.

THEOREM' XIII.

150. The perpendicular is the least sect between a given point and a given line.

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HYPOTHESIS. Let A be a given point, and BC a given line.
CONSTRUCTION. By 139, from A drop ADI BC.
By 100, join A to F, any point of BC except D.
CONCLUSION. AD< AF.
PROOF. Since, by construction, 4 ADF is rt. 4,

.. 4 ADF > * AFD,

(143. Any two angles of a triangle are together less than a straight angle.)

.:. AF > AD.

(148. If angle ADF is greater than angle AFD, therefore side AF is greater than

side AD.)

151. Except the perpendicular, any sect from a point to a line is called an Oblique.

EXERCISES. 14. Determine the least path from one point to another, subject to the condition that it shall meet a given straight line.

15. The four sides of a quadrilateral figure are together greater than its two diagonals.

THEOREM XIV.

152. Two obliques from a point, making equal sects from the foot of the berpendicular, are equal, and make equal angles with the line.

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(124. Triangles are congruent if two sides and the included angle are equal in each.)

153. COROLLARY. For every oblique, there can be drawn one equal, and on the other side of the perpendicular.

EXERCISES. 16. ABC is a triangle whose angle A is bisected by a line meeting BC at D. Prove AB greater than BD, and AC greater than CD.

17. Prove that the sum of the sects from any point to the three angles of a triangle is greater than one-half the perimeter of the triangle.

18. The two sides of a triangle are together greater than twice the line drawn from the vertical angle to the middle point of the base.

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