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cumference of A does not pass through the point F; and the radius GK being greater than GF, the point F is within the circle A. Now (by Hyp.) the circles intersect in C; and it has been proved that every point in the circumference of the circle B, on the one side of the point D, is without the circle A, and every point on the other side within it; therefore D is the only other point common to the circumferences of the circles A and B, or in which they cut one another.
To construct a triangle of which the three sides shall be equal to three given straight lines, but any two whatever of these must (Prop. 8) be greater than the third.
Let A, B, and C be three given straight lines, of which any two whatever are greater than the third, it is required to make a triangle of which the three sides shall be equal to A, B, and C, each to each.
From the point D draw (Prop. I.) the straight line DE, equal to B; and from D as a centre, with a radiant equal to A, describe (Post. 3) the
K circle FGH; and from E as a centre, with a radiant equal to C, describe the circle FGK. Now if A and C were
A less than B, the one circle must evidently lie wholly without the other; if A and C were equal to B, the circles must meet in some point of B or DE; but if A and C are greater than B, the one circle would lie partly within and partly without the other, and the circles must therefore intersect. Let F be the point of intersection, and join FD and FE, FDE is the triangle required.
For DF is a radius of the circle FGH, and therefore equal
(Def. 34, Cor.) to the radiant; but the radiant was taken equal to A; therefore (Ax. 1) DF is equal to A. For the same reason, FE is equal to C, and, by construction, DE is equal to B; therefore the triangle FDE has its three sides equal to the three given straight lines A, B, and C, each to each. It is also evident, that if the lines A and C were equal, the triangle DEF would be isosceles; and if these were also equal to B, the triangle DEF would be equilateral. Observe, also, that if straight lines were drawn from D and E to the point G, a triangle would be formed on the other side of DE, having its sides respectively equal to the same three given straight lines.
Cor. On the same base, and on the same side of it, there cannot be two triangles which have their two sides which are tern
nated in one extremity of the base equal another, and likewise those which are terminated in the other extremity equal to one another; for since the two circles cannot cut (9) each other in more points than F and G, it is evident that DEF is the only triangle that can be formed on the same side of the base DE, having its two sides respectively equal to A and C.
Scholium. In practice, it is only necessary to describe two arcs in order to determine the position of the point F, and then draw the straight lines FD and FE.
If the three sides of one triangle are equal to the three sides of another, each to each, the triangles are equal, and have those angles equal which are opposite to the equal sides.
Let ABC and DEF be two triangles which have the side AB equal to DE, AC to DF, and BC to EF, the triangles are equal (Def. 37).
For if the triangle ABC be applied to the triangle DEF,
so that the point B be on E, and the straight line BC on EF, the
G point C shall coincide with the point F, because BC is equal to EF; the side BC therefore coinciding with EF, the sides BA B and AC shall coincide with DE and EF; for if BA and AC do not coincide with ED and DF, then upon the same base and on the same side of it there can be two triangles, EDF and EGF, that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity; but this is (10, Cor.) impossible; therefore if the base BC coincide with EF, the sides BA, AC, cannot but coincide with ED and DF, and the point A shall coincide with D. Now the two plane surfaces ABC and DEF have (Hyp.) equal boundaries, and coinciding, as proved, in three points not in the same straight line, they must (Def. 6, Cor.) coincide in every part, and be in all respects equal (Def. 37): their areas are equal, the angle A is equal to D, the angle B to DEF, and C to DFE, the equal angles being always opposite to the equal sides.
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 triangles are equal (Def. 37).
Let the two triangles ABC and EDF have the side AB equal to DE, AC to DF, and the included angle A equal to D, the base BC shall be equal to the base EF, and the triangles shall be every way equal.
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 on DE, the point B shall coincide with the point E, because AB is equal
to DE; and AC shall coincide
A with DF, because the angle A is equal to D; and the point C shall coincide with F, because AC is equal to DF. But the point B
CE coincides with E; wherefore the base BC shall coincide (Def. 3, Cor. 1) with the base EF, and be equal to it. Since therefore the two plane surfaces ABC and DEF have equal boundaries, and coincide in three points not in the same straight line, they must coincide (Def. 6, Cor.) in every part, and be in all respects equal; their areas are equal, the angle B is equal to the angle E, and the angle C to F.
To bisect a given rectilineal angle.
Let BAC be a given rectilineal angle, it is required to bisect it, that is, to divide into two equal parts. Take any point D in AB, and from AC cut
A off (2) AE, equal to AD; join DE, and on DE describe (10) the isosceles triangle DEF; join AF; the straight line AF bisects the
E angle BAC. Because in the triangles ADF and AEF, the side AD is equal to AE, the
B side FD to FE, and AF is common to both
F triangles; therefore (11) the triangles are equal; wherefore the angle DAF is equal to the angle EAF, and the angle BAC is bisected by the straight line AF.
If two sides of a triangle are equal, the angles opposite to them are also equal.
In the triangle ABC, let the side AB be equal to the side AC, then the angle B is also equal to the angle C.
Bisect the angle BAC (13) by the straight line AD, then, in the two triangles ABD and ACD, the side AB is equal to the side AC; the side AD is common to both, and the included angles BAD and CAD are equal; hence (12) the triangles are equal, and therefore the angle B is equal to the angle C.
Cor. 1. Hence every equilateral triangle is also equiangular.
Cor. 2. The straight line bisecting the vertical angle of an isosceles triangle bisects also the base at right angles; for the adjacent angles ADB and ADC are equal, and BD is equal DC.
Scholium. The enunciation of this proposition, given by Euclid and many other geometers, is—“The
gles at the base of an isosceles triangle are equal."
To bisect a given finite straight line.
Let AB be a given finite straight line, it is required to bisect it.
On AB construct (10) two isosceles triangles ACB and ADB, and join their vertices by the straight line CD, cutting AB in E, the line AB is bisected in E.
For in the triangles ACD and BCD the side A AC is equal to BC, the side AD to BD, and the side CD is common to both, the triangles are D therefore (11) equal, and the angle ACD or ACE is equal to BCD or BCE; but the straight line CE which bisects the vertical angle of the isosceles triangle ACB, bisects (14, Cor. 2), also its base at right angles; therefore AE is equal EB, and the straight line AB is bisected in E.