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PROPOSITION I. THEOREM.
If one straight line meet another straight line, the sum of the two adjacent angles will be equal to two right angles.
Let the straight line DC meet the straight line AB at C, then will the angle ACD + the angle DCB, be equal to two right angles.
At the point C, erect CE perpendicular to AB. The angle ACD is the sum of the an-gles ACE, ECD: therefore ACD+DCB is the sum of the three angles ACE, ECD, DCB: but the first of these three angles is a right angle, and the other two make up the right angle ECB; hence, the sum of the two angles ACD and DCB, is equal to two right angles.
Cor. 1. If one of the angles ACD, DCB, is a right angle, the other must be a right angle also.
Cor. 2. If the line DE is perpendicular to AB, reciprocally, AB will be perpendicular to DE.
For, since DE is perpendicular to AB, the A angle ACD must be equal to its adjacent angle DCB, and both of them must be right angles (Def. 10.). But since ACD is a
right angle, its adjacent angle ACE must also be a right angle (Cor. 1.). Hence the angle ACD is equal to the angle ACE, (Ax. 10.): therefore AB is perpendicular to DE.
Cor. 3. The sum of all the successive angles, BAC, CAD, DAE, EAF, formed on the same side of the straight line BF, is equal to two right angles; for their sum is equal to that of the two adjacent an-, *gles, BAC, CAF.
PROPOSITION II. THEOREM.
Two straight lines, which have two points common, coincide with each other throughout their whole extent, and form one ana the same straight line.
Let A and B be the two common points. In the first place it is evident that the two lines must coincide entirely between A and B, for otherwise there would be two straight lines between A and B, which is impossible (Ax. 11). Sup
pose, however, that on being produced, these lines begin to separate at C, the one becoming CD, the other CE. From the point C draw the line CF, making with AC the right angle ACF. Now, since ACD is a straight line, the angle FCD will be a right angle (Prop. I. Cor. 1.); and since ACE is a straight line, the angle FCE will likewise be a right angle. Hence, the angle FCD is equal to the angel FCE (Ax. 10.); which can only be the case when the lines CD and CE coincide: there fore, the straight lines which have two points A and B common, cannot separate at any point, when produced; hence they form one and the same straight line.
PROPOSITION III. THEOREM.
If a straight line meet two other straight lines at a common point, making the sum of the two adjacent angles equal to two right angles, the two straight lines which are met, will form one and the same straight line.
Let the straight line CD meet the two lines AC, CB, at their common point C, making the sum of the two adjacent angles DCA, DCB, equal to A two right angles; then will CB be the prolongation of AC, or AC and CB will form one and the same straight line.
For, if CB is not the prolongation of AC, let CE be that prolongation: then the line ACE being straight, the sum of the angles ACD, DCE, will be equal to two right angles (Prop. I.). But by hypothesis, the sum of the angles ACD, DCB, is also equal to two right angles: therefore, ACD+DCE must be equal to ACD+DCB; and taking away the angle ACD from each, there remains the angle DCE equal to the angle DCB, which can only be the case when the lines CE and CB coincide ; hence, AC, CB, form one and the same straight line.
When two straight lines intersect each other, the opposite or vertical angles, which they form, are equal.
Let AB and DE be two straight A lines, intersecting each other at C; then will the angle ECB be equal to the angle ACD, and the angle ACE to the angle DCB.
For, since the straight line DE is met by the straight line AC, the sum of the angles ACE, ACD, is equal to two right angles (Prop. I.); and since the straight line AB, is met by the straight line EC, the sum of the angles ACE and ECB, is equal to two right angles: hence the sum ACE+ACD is equal to the sum ACE+ECB (Ax. 1.). Take away from both, the common angle ACE, there remains the angle ACD, equal to its opposite or vertical angle ECB (Ax. 3.).
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 sum of the two angles ACE, ECB, is equal to two right angles; and the sum of the other two, ACD, DCB, is also equal to two right angles: therefore, the sum of the four is equal to four right angles.
In general, if any number of straight lines CA, CB, CD, &c. meet in a point C, the sum of all the successive angles ACB, BCD, DCE, ECF, FCA, will be equal to four right angles: for, if four right angles were formed about the point C, by two lines pendicular to each other, the same space would be occupied by the four right angles, as by the successive angles ACB, BCD, DCE, ECF, FCA.
PROPOSITION V. THEOREM.
If two triangles have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the two triangles will be equal.
For, these triangles may be so applied to each other, that they shall exactly coincide. Let the triangle EDF, be placed upon the triangle BAC, so that the point E shall fall upon B, and the side ED on the equal side BA; then, since the angle D is equal to the angle A, the side DF will take the direction AC. But
DF is equal to AC; therefore, the point F will fall on C, and the third side EF, will coincide with the third side BC (Ax. 11.): therefore, the triangle EDF is equal to the triangle BAC (Ax. 13.).
Cor. When two triangles have these three things equal, namely, the side ED=BA, the side DF=AC, and the angle D=A, the remaining three are also respectively equal, namely, the side EF=BC, the angle E=B, and the angle F=C
PROPOSITION VI. THEOREM.
If two triangles have two angles and the included side of the one, equal to two angles and the included side of the other, each to each, the two triangles will be equal.
Let the angle E be equal to the angle B, the angle F to the angle C, and the included side EF to the included side BC; then will the triangle EDF be equal to the triangle BAC.
For to apply the one to the other, let the side EF be placed on its equal BC, the point E falling on B, and the point F on C; then, since the angle E is equal to the angle B, the side ED will take the direction BA; and hence 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, falling at the same time in the two straight lines BA and CA, must fall at their intersection A: hence, the two triangles EDF, BAC, coincide with each other, and are therefore equal (Ax. 13.).
Cor. Whenever, in two triangles, these three things are equal, namely, the angle E=B, the angle F-C, and the included side EF equal to the included side BC, it may be inferred that the remaining three are also respectively equal, namely, the angle D=A, the side ED=BA, and the side DF-AC.
Scholium. Two triangles are said to be equal, when being applied to each other, they will exactly coincide (Ax. 13.). Hence, equal triangles have their like parts equal, each to each, since those parts must coincide with each other. The converse of this proposition is also true, namely, that two triangles which have all the parts of the one equal to the parts of the other, each
to each, are equal; for they may be applied to each other, and the equal parts will mutually coincide.
PROPOSITION VII. THEOREM.
The sum of any two sides of a triangle, is greater than the third side.
Let ABC be a triangle: then will the sum of two of its sides, as AC, CB, be greater than the third side AB.
For the straight line AB is the shortest distance between the points A and B (Def. 3.); hence AC+CB is greater A than AB.
PROPOSITION VIII. THEOREM.
If from any point within a triangle, two straight lines be drawn to the extremities of either side, their sum will be less than the sum of the two other sides of the triangle.
Let any point, as O, be taken within the triangle BAC, and let the lines OB, OC, be drawn to the extremities of either side, as BC; then will OB+OC<BA+AC.
Let BO be produced till it meets the side AC in D: then the line OC is shorter than OD+DCB
(Prop. VII.): add BO to each, and we have BO+OC<BO+ OD+DC (Ax. 4.), or BO+OC<BD+DC.
Again, BD<BA+AD: add DC to each, and we have BD+ DC BA+AC. But it has just been found that BO+OC< BD+DC; therefore, still more is BO+OC<BA+AC.
PROPOSITION IX. THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides will be unequal; and the greater side will belong to the triangle which has the greater included angle.
Let BAC and EDF be two triangles, having the side AB=DE, AC =DF, and the angle A>D; then will BC> EF.
Make the angle CAGB4 =D; take AG=DE, and draw CG.