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PROPOSITION I. THEOREM.
If two triangles have two sides and the included angle in the one, equal to two sides and the included angle in the other, the triangles will be identical, or equal in all respects.
Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and let the angle BAC be also equal to the angle EDF: then shall the base BC be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles, to which the equal sides are opposite, shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB. to DFE.
For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because AC is equal to DF: But the point B coincides with the point E; wherefore the base BC shall coincide with the base EF (cor. def. 3.), and shall be equal to it. Therefore also the whole triangle ABC shall coincide with the whole triangle DEF, so that the spaces which they contain or their areas are equal; and the remaining angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.
PROP. II. THEOR.
When two triangles have two angles and the interjacent sides in the one, equal to two angles and the interjacent sides in the other, the triangles are identical, or have their other sides and angles equal.
Let the two triangles BAC, EDF, (see the last figure,) have the angle ual to the angle E, the angle C equal to the angle F, and the side BC equal to the side EF; then these two triangles will be identical.
For, conceive the triangle BAC to be placed on the triangle EDF, in such manner that the side BC may fall exactly on the equal side EF. Then, since the angle B is equal to the angle E (by hyp.) the side BA must fall on the side ED; and, in like manner, because the angle C is equal to the angle F, the side AC must fall on the side DF. Hence, the point A,
occurring at the same time in the two straight lines ED and FD, inust fall on their intersection D; consequently, the two triangles BAC, EDF, are identical, or mutually equal (Ax. 8.).
COROLLARY. Whenever in two triangles these three things are equal, namely, BC=EF, B=E, and C=F, it may be inferred that the other three are equal also; that is, AB=DE, AC=DF, and the angle BAC= DEF.
PROP. III. THEOR.
The angles at the base of an isosceles triangle are equal to one another : namely, those to which the equal sides are opposite.
If the triangle ABC have the side AC equal to the side BC: then will the angle B be equal to the angle A.
For, conceive the angle C to be bisected, or divided into two equal parts by the line CD, making the angle ACD equal to the angle BCD.
Then the two triangles ACD, BCD have two sides and the contained angle of the one equal to two sides and the contained angle of the other; namely, the side AC equal to BC, the angle ACD equal to BCD, and the side CD common; therefore, these two triangles are identical, or equal in all respects, (Theor. 1.); and consequently, the angle A equal to the angle B.
COR. Every equilateral triangle is also equiangular.
The equality of the triangles ADC, BDC, proves also that AD is equal to BD, and the angle ADC equal to the angle BDC; hence, these two are right angles: hence, the line bisecting the vertical angle of an isosceles triangle is perpendicular to its base, and divides that base into two equal parts.
In a triangle which is not isosceles, any side may be assumed indifferently as the base; and the vertex is, in that case, the vertex of the opposite angle. In an isosceles triangle, however, that side is specially assumed as the base, which is not equal to either of the other two.
PROP. IV. THEOR.
two angles of a triangle be equal to one another, the sides which subtend, or are opposite to them, are also equal to one another.
Let ABC be a triangle having the angle ABC equal to the angle ACB; the side AB is also equal to the side AC.
For, if AB be not equal to AC, one of them is greater than the other: Let AB be the greater, and from it cut off DB equal to AC the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; but the angle DBC is also equal to the angle ACB; therefore the base DC is equal to the base AB, and the area of the triangle DBC is equal to that of the triangle (1. 1.) ACB, the less to the greater; which is absurd. Therefore, AB is not unequal to AC, that is, it is equal to it.
COR. Hence every equiangular triangle is also equilateral.
PROP. V. THEOR.
When two triangles have all the three sides in the one, equal to all the three sides in the other, the triangles are identical, or have also their three angles equal, each to each.
Let the two triangles ABC, DEF, have their three sides respectively equal, that is, the side AB equal to DE, AC to DF, and BC to EF; then shall the two triangles be identical, or have their angles equal, that is, those angles that are opposite to the equal sides; namely, the angle BAC to the angle EDF, the angle ABC to DEF, and the angle ACB to DFE.
For, conceive the two triangles to be joined together by their longest equal sides; that is, having GB equal to DE, GC to DF; and draw the line AG.
Then, in the triangle ABG, because the side AB is equal to BG or DE, (by hyp.) the angle BAG is equal to the angle BGA, (Th. 3.). In like manner, in the triangle ACG, the angle CAG is equal to the angle CGA, because the side AC is equal to CG or DF. Hence, then, the angle BAG
being equal to the angle BGA, and the angle CAG equal to the angle CGA, by equal additions, the sum of the two angles BAG, CAG, is equal to the sum of the two angles BGA, CGA, (Ax. 2.) that is, the whole angle BAC equal to the whole angle BGC.
Since, then, the two sides BA, AC, are equal to the two sides BG, GC, each to each, and their contained angles BAC, BGC, also equal, the two triangles BAC, BGC or EDF, are identical (Th. 1.) and have the other angles equal; namely, the angle ABC equal to GBC, and the angle ACB equal to the angle GCB.
PROP. VI. THEOR.
The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles.
Let the straight line AB make with CD, upon one side of it the angles CBA, ABD; these are either two right angles, or are together equal to two right angles.
For, if the angle CBA be equal to ABD, each of them is a right angle (Def. 7.); but, if not, let BE be perpendicular to DC at the point B. The angle DBA is the sum of the angles DBE, EBA: therefore, DBA+ABC is the sum of the three angles DBE, EBA, ABC: but the first of those three angles is a right angle; and the other two together make up the right angle CBE; hence the sum of the two angles DBA and ABC is equal to two right angles.
COR. The sum of all the angles, formed on the same side of a straight line DC, is equal to two right angles; because their sum is equal to that of the two adjacent angles DBA, ABC.
PROP. VII. THEOR.
If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line.
At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the same straight line with CB.
For if BD be not in the same straight line with CB, let BE be in the same straight line with it; therefore, because the straight line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal (6. 1.) to two right angles; but the angles ABC, ABD are likewise together equal to two right angles therefore the angles CBA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, and the remaining angle ABE is equal (3. Ax.) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB.
PROP. VIII. THEOR.
If two straight lines cut one another, the vertical, or opposite angles are equal. Let the two straight lines AB, CD, cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED.
For the angles CEA, AED, which the straight line AE makes with the straight line CD, are together equal (6. 1.) to two right angles and the angles AED, DEB, which the straight line DE makes with the straight line AB, are also together equal (6. 1.) to two right angles; two AED, DEB. ing angle CEA is
therefore the two angles CEA, AED are equal to the Take away the common angle AED, and the remainequal (3. Áx.) to the remaining angle DEB. In the same manner it may be demonstrated that the angles CEB, AED are equal.
COR. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles.
COR. 2. And hence, all the angles made by any number of straight lines meeting in one point, are together equal to four right angles.
PROP. IX. THEOR.
If one side of a triangle be produced, the exterior angle is greater than either of the interior, and opposite angles.
Let ABC be a triangle, and let its side BC be produced to D, the exte