angle ABH is equal to the angle DEF [Hyp.]; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite [I. 4]; therefore the angle BHA is equal to the angle EFD. But the angle EFD is equal to the angle BCA. [Hyp.] Therefore the angle BHA is equal to the angle BCA [Ax. 1]; that is, the exterior angle BHA of the triangle AHC is equal to the interior and opposite angle BCA; which is impossible. [I. 16.] Therefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE [Hyp.]; therefore the two sides AB, BC are equal to the two sides DE, EF, each to each; and the angle ABH is equal to the angle DEF [Hyp.]; therefore, the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. [I. 4.] Therefore, if two triangles, &c. Q.E.D. 1 It is evident that the triangles are equal in all respects, as in props. 4 and 8. This prop. closes the first section of the first Book of the "Elements." It may be said to treat of Triangles, and the results established show:-1. The conditions under which Triangles are equal to one another. 2. The conditions under which the sides and angles of triangles are either equal or unequal to one another. The next section of the book extends to prop. 34, and treats of the properties of parallel lines. The following construction has important consequences, and may be introduced here:- To construct a triangle, having given two sides and an angle opposite to one of them. Let AB and BH be the two given sides, and GAH the given angle. It is required to construct a triangle whose sides shall be equal to AB and BH, and having an angle equal to BAH. With centre B and radius BH describe a circle. 1. If this circle do not meet AG, there is no triangle with the required parts. 2. If this circle meet AG in one point only, there is one, and only one, triangle having the required parts. 3. In general the circle will cut AG in two points, as in the fig., viz., at D and C. Join DB, CB, and we thus obtain two triangles with the required parts, viz., ADB, ACB. And it should be noticed that there is this difference between them: the angle ACB opposite AB in the one triangle is acute (for DCB is an isosceles triangle), while ADB, the angle in the other triangle opposite to AB, is obtuse. Therefore, when two triangles have two sides of the one equal to two sides of the other, each to each, and the angle opposite to one pair of equal sides, equal in each triangle, the triangles will not be equal unless the angle opposite the other pair of equal sides is of the same species in each triangle, ¿.e., acute in both triangles, or obtuse in both triangles. But if this latter condition be complied with, the triangle is determinate. Ex. 1. Any point in the line bisecting an angle is equidistant from the two sides containing the angle. 2. If the straight line which bisects the vertical angle of a triangle cuts the base at right angles, the triangle is isosceles. PROPOSITION 27. THEOREM. If a straight line falling on two other straight lines, make the alternate angles equal to one another, the two straight lines shall be parallel to one another. Let the straight line EF, which falls on the two A 1. straight lines AB, CD, make the alternate angles AEF, EFD equal to one another: AB shall be parallel to CD. B For if it be not parallel, AB and CD being pro2. duced, will meet either towards B, D, or towards A, D. Let them be produced and meet towards B, D in the point G. 3. Then because GEF is a triangle, its exterior angle AEF is greater than the interior and opposite angle EFG. [I. 16.] But the angle AEF is also equal to the angle EFG [Hyp.]; which is impossible. Therefore AB and CD being produced, do not meet towards B, D. In like manner, it may be proved that they do not meet towards A, C. But those straight lines which being produced ever so far both ways do not meet, are parallel. [Def. 35.] Therefore AB is parallel to CD. Therefore, if a straight line, &c. Q.E.D PROPOSITION 28. THEOREM. If a straight line falling on two other straight lines, make the exterior angle equal to the interior and opposite angle on the same side of the line, or make the interior angles on the same side together equal to two right angles, the two straight lines shall be parallel to one another. 1. Let the straight line EF, which falls on the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD on the same side: AB shall be parallel to CD. Because \E 3. the angle EGB is equal to the angle AGH [I. 15], therefore A the angle AGH is equal to the angle GHD [Ax. 1], and they are alternate angles; therefore AB is parallel to CD. [I. 27.] 1. Let the straight line EF fall on the two straight lines AB, CD, and make the interior angles BGH, GHD on the same side together equal to two right angles: AB shall be parallel to CD. Then because the angles BGH, GHD are to3. gether equal to two right angles [Hyp.], and that the angles AGH, BGH are also together equal to two right angles [I. 13]; therefore the angles AGH, BGH are equal to the angles BGH, GHD. Take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHI [Ax. 3]; and they are alternate angles; therefore AB is parallel to CD. [I. 27.] Therefore, if a straight line, &e. Q.E.D. Propositions 27 and 28 show that under certain conditions lines are parallel to each other. These conditions are: 1. That the alternate angles are equal; or (2), that the exterior angle is equal to the interior and opposite; or (3), that the two interior angles are together equal to two right angles; and from the demonstrations it is evident that all these conditions amount to the same thing. But if these conditions do not exist, may the lines still be parallel? Suppose, for example, the two interior angles are together less than two right angles will the lines be parallel, or will they meet if produced? In other words, can there be two different lines passing through the same point, both of which are parallel to a third line? These questions, neither the definition of parallel lines nor the propositions already established, enable us to answer. But until they are answered our knowledge of parallel lines is evidently incomplete; and we are, therefore, driven to one of two courses, viz.: 1st. To make a definition which shall tell us something more of parallel lines than that they never meet; or, 2nd. To assume some other property of such lines, i.e., to lay down a new axiom. Euclid adopted the latter course, and introduced his famous 12th Axiom. This method of supplying the defect is probably as simple as any other when the axiom is thus stated. "C Two straight lines which cut one another cannot be both parallel to a third straight line.” With the help of this axiom we can now show that all parallel lines fulfil the conditions above laid down, and therefore that such lines, and such only, are parallel to one another. PROPOSITION 29. THEOREM. If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles. 1. Let the straight line EF fall on the two parallel straight lines AB, CD: the alternate angles AGH, GHD shall be equal to one another, and the exterior angle EGB shall be equal to the interior and opposite angle on the same side, GHD, and the two interior angles on the same side, BGH, GHD, shall be together equal to two right angles. E 2. For if the A. Κ be not equal to the angle GHD, one of them must be greater than the other; let the angle AGH be C the greater. Then since AGH is greater than GHD, at the point Ginthe straight F line GH make the angle KGH equal to the angle GHD. [I. 23.] Produce KG to L. H B 3. Then since GH falls upon the two straight lines KL and CD, and makes the angle KGH equal to the alternate angle GHD; [Const.] therefore KL is parallel to CD. But AB is parallel to CD. [Hyp.] Therefore KL and AB, which cut one another, are both parallel to CD. But this is impossible. [Ax. 12.] Therefore the angle AGH is not unequal to the angle GHD; that is, it is equal to it. But the angle AGH is equal to the angle EGB [I. 15.] Therefore the angle EGB is equal to the angle |