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ACE+BCE. (Ax. 1.) Take away from both the same angle ACE; there remains the angle ACD, equal to its opposite or vertical angle BCE. (Ax. 3.)
It may be shown, in the same manner, that the angle ACE is equal to its opposite angle BCD. Hence, universally,
If any two straight lines intersect each other, the opposite or vertical angles are equal.
Scholium. The four angles formed about a point by two straight lines which intersect each other, are together equal to four right-angles: for the two angles ACE, BCE, taken together, are equal to two right-angles; and the other two, ACD, BCD, are likewise equal to two right-angles.
In general, if any number of straight lines CA, CB, CD, &c. meet in a point C, the sum of all A the angles ACB, BCD, DCE, ECF, FCA, thus formed, will be equal to four right-angles: for if four right-angles were formed about F
the point C, by means of two lines
perpendicular to each other, they would occupy the same space as the successive angles ACB, BCD, DCE, ECF, FCA.
PROPOSITION V. THEOREM.
If two triangles ABC, DEF, have two sides and the included angle of the one respectively equal to two sides and the included angle of the other, viz. the side AB to DE, AC to DF, and the angle A to the angle D, the triangles are every way equal.
and, since the angle D is equal to the angle A, when the side DE is placed on AB, the side DF will take the direction AC. But by hypothesis, DF is equal to AC; therefore the point F will fall on C, and the third side EF must exactly coincide with the third side BC; (Ax. 12;) therefore the triangle DEF coincides with the triangle ABC throughout, consequently they are equal. (Ax. 13.) Hence, universally,
Any two triangles are every way equal, when two sides and the included angle of the one, are respectively equal to two sides and the included angle of the other.
If two triangles ABC, DEF, have two angles and the interjacent side of the one respectively equal to two angles and the interjacent side of the other; viz., the angle B to the angle E, the angle C to F, and the side BC to EF, the two triangles are every way equal.
For, if the side
EF be placed on its equal BC, the point E will fall
on B, and the point F on C.
And, since the angle E is equal to the angle B, the side ED will take the direction BA; therefore, the point D will be found somewhere in the line BA. In like manner, since the angle F is equal to the angle C, the line FD will take the
direction CA, and the point D will be found somewhere in the line CA. Hence, the point D, occurring at the same time in the two straight lines BA and CA, must fall on their intersection A; hence the two triangles ABC, DEF, coincide with each other, and are consequently equal. (Ax. 13.) Hence, universally,
Any two triangles are every way equal, when two angles and the interjacent side of the one, are respectively equal to two angles and the interjacent side of the other.
In every triangle, any side is less than the sum of the other two.
For the straight line BC, for example,
is the shortest distance from B to C; (Def. 5 ;) therefore, BC is less than BA+ AC. Hence, In every triangle, &c.
If from a point O within the triangle ABC, straight lines OB, OC, be drawn to the extremities of a side BC, the sum of these straight lines will be less than that of the two other sides AB, AC.
Let BO be produced till it meet the side AC in D. The line OC, is shorter than OD+DC: (Prop. 7:) add BO to each, and we have BO+OC<BO+0D+DC,, (Ax. 4,) or BO+OC<BD+DC.
Moreover, BD<BA+AD: add DC
to each; and we have BD+DC<BA+AC. But we have just found BO+OC<BD+DC; therefore, still more is BO+OC<BA+AC.
In like manner, if from the point O, straight lines be drawn to the extremities of the side AB, or AC, it may be proved that AO+BO<AC+BC; and AO+CO<AB+BC. Hence,
If from any point within a triangle, two straight lines be drawn to the extremities of either side, the sum of these two lines will be less than the sum of the other two sides of the triangle.
PROPOSITION IX. THEOREM.
If the two sides AB, AC, of the triangle ABC, are respectively equal to the two sides DE, DF, of the triangle DEF, while, at the same time, the angle BAC contained by the former, is greater than EDF contained by the latter; then will the third side, BC of the first triangle, be greater than EF, the third side of the second.
each, contained by equal sides; therefore CG is equal to EF. Now, there may be three cases in the proposition, according as the point G falls without the triangle ABC, or upon its base BC, or within it.
First. The straight line GC is shorter than GI+IC, and the straight line AB is shorter than AI+IB; (Prop.7;) therefore, GC+AB is shorter than GI+AI+IC+IB, or, which is the same thing, GC+AB<AG+BC. Take AB
from the one side, and its equal AG from the other, there will remain GC<BC; (Ax. 5;) but we have found GC= EF, therefore, EF<BC.
Third. If the point G fall within the triangle ABC, we shall have, AG+GC<AB+BC; (Prop. 8;) and, taking AG from one side, and its equal B AB from the other, there will remain GC<BC, or EF<BC. Hence,
If two triangles have two sides of the one respectively equal to two sides of the other, while the included angle of the one
is greater than the included angle of the other, then will the third side of that which has the greater angle be greater than the third side of the other.
Scholium. Conversely, if the two sides AB, AC of the triangle ABC are respectively equal to the two DE, DF of the triangle DEF, while the third side, CB of the first triangle, is greater than the third EF of the second; then will the angle BAC of the first triangle be greater than the angle EDF of the second.
For if not, the angle BAC must be equal to, or less than EDF. If equal, the side CB would be equal to EF. (Prop. 5.) If less, CB would be less than EF: but both of these results contradict the hypothesis; therefore, BAC is greater than EDF.
PROPOSITION X. THEOREM.
If two triangles ABC, DEF, have the three sides of the one respectively equal to the three sides of the other, viz. AB to