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Or, since AG = AB, and
AB, and GC =
GC = EF, we have,
AB + BC > AB + EF.
Taking away the common part AB, there remains (A. 5),
BC > EF.
2°. When G is on BC.
In this case, it is obvious that GC is less than BC; or since GC = EF, we have,
BC > EF.
3°. When G is within the triangle ABC. From Proposition VIII., we have,
Taking away the common part AB, there remains,
F BC > EF. Hence, in each case, BC is greater than EF; which was to be proved.
Conversely: If in two triangles ABC and DEF, the side AB is equal to the side DE, the side AC to DF, and BC greater than EF, then is the angle BAC greater than the angle EDF.
For, if not, BAC must either be equal to, or less than, EDF. In the former case, BC would be equal to EF
, (P. V.), and in the latter case, BC would be less than EF; either of which
which would contradict the hypothesis : hence, BAC must be greater than EDF.
If two triangles have the three sides of the one equal to
the three sides of the other, each to each, the triangles are equal in all respects.
In the triangles ABC and DEF, let AB be equal to DE, AC to DF, and BC to EF: then are the triangles equal in all respects.
For, since the sides AB, AC, are equal to DE, DF, each to each, if the angle A were greater than D, it would follow, by the last Proposition, that the side BC would be greater than EF; and if the
с Е angle A were less than D, the side BC would be less than EF. But BC is equal to EF, by hypothesis ; therefore, the angle A can neither be greater nor less than D: hence, it must be equal to it. The two triangles have, therefore, two sides and the included angle of the one equal to two sides and the included angle of the other, each to each; and, consequently, they are equal in all respects (P. V.); which was to be proved.
Scholium. In triangles, equal in all respects, the equal sides lie opposite the equal angles; and conversely.
In an isosceles triangle the angles opposite the equal sides
Let BAC be an isosceles triangle, having the side AB equal to the side AC; then the angle C is equal to the angle B.
Join the vertex A and the middle point D of the base BC.
Then, AB is equal to AC, by hypothesis, AD common, and BD equal to DC, by construction: hence, the triangles BAD, and DAC, have the three sides of the one equal to those of the other, each to each; therefore, by the last Proposition, the angle B is equal to the angle C; which was to be proved.
Cor. 1. An equilateral triangle is equiangular.
Cor. 2. The angle BAD is equal to DAC, and BDA to CDA: hence, the last two are right angles. Consequently, a straight line drawn from the vertex of an isosceles triangle to the middle of the base, bisects the angle at the vertex, and is perpendicular to the base.
If two angles of a triangle are equal, the sides opposite to
them are also equal, and consequently, the triangle is isosceles.
In the triangle ABC, let the angle ABC be equal to the angle ACB: then is AC equal to AB, and consequently, the triangle is isosceles.
For, if AB and AC are not equal, suppose one of them, 23 AB, to be the greater. On this, take BD equal to AC (Post. 3), and draw DC. Then, in the triangles ABC, DBC, we have the side BD equal to AC, by construction, the side BC common, and the included angle ACB equal to the included angle DBC, by hypothesis: hence, the two triangles are equal
in all respects (P. V.). But this is impossible, because a part can not be equal to the whole (A. 8): hence, the hypothesis that AB and AC are unequal, is false. They must, therefore, be equal; which was to be proved.
An equiangular triangle is equilateral.
In any triangle, the greater side is opposite the greater
angle; and, conversely, the greater angle is opposite the greater side.
In the triangle ABC, let the angle ACB be greater than the angle ABC: then the side AB is greater than the side AC.
For, draw CD, making the angle BCD equal to the angle B (Post. 7): then, in the triangle DCB, we have the angles DCB and DBC equal: hence, the opposite sides DB and DC are equal (P. XII.). In the triangle ACD, we have (P. VII.),
AD + DC > AC;
or, since DC = DB, and AD + DB = AB, we have,
AB > AC;
which was to be proved.
Conversely: Let AB be greater than AC: then the angle ACB is greater than the angle ABC.
For, if ACB were less than ABC, the side AB would be less than the side AC, from what has just been proved ; if ACB were equal to ABC, the side AB would be equal to AC, by Prop. XII.; but both conclusions contradict the hypothesis: hence, ACB can neither be less than, nor equal to, ABC; it must, therefore, be greater; which was to be proved.
From a given point only one perpendicular can be drawn
to a given straight line.
Let A be a given point, and AB
A a perpendicular to DE: then can other perpendicular to DE be drawn
E from A.
For, suppose a second perpendicular AC to be drawn. Prolong AB till BF is
F equal to AB, and draw CF. Then, the triangles ABC and FBC have AB equal to BF, by construction, CB common, and the included angles ABC and FBC equal, because both are right angles: hence, the angles ACB and FCB are equal (P. V.). But ACB is, by a hypothesis, a right angle: hence, FCB must also be a right angle, and consequently, the line ACF must be a straight line (P. IV.). But this is impossible (A. 11). The hypothesis that two perpendiculars can be drawn is, therefore, absurd; consequently, only one such perpendicular can be drawn; which was to be proved.
If the given point is on the given line, the proposition is equally true. For, if from A two perpendiculars AB and AC could be drawn to DE, we
B should have BAE and CAE each equal to a right angle; and consequently, equal to each other; which is absurd
-E (A. 8).