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EU. I. 8. THEOR.
If two triangles have two sides of the one equal to two sides of the other, each to each, and have also their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them of the other. In triangles ABC and DEF,
A let AB = DE, AC = DF,
and BC = EF. To
prove LA = _D. Apply A ABC to A DEF so that B is on E and
falls on F, ::: BC = EF (given). And if ABAC do not coincide with AEDF, let it fall as
AEGF. Then :: EG = BA, and BA = ED, ... ED = EG; and in like manner FG=FD; which is impossible (I. 7). .:: ABAC must coincide with AEDF,
.. LA = LD. Q.E.D.
COR.—AABC coincides with ADEF, and .:. AABC
ADEF in every respect.
EU. I. 24. THEOR.
If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides, equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other.
Let triangles ABC and DEF
D have AB = DE, AC = DF,
and LA> LEDF. Το prove
BC “Let DE be the side which
E is not greater than the B
other.” At D in ED make EDG = _BAC (I. 23), and make DG DF , AC (1. 3). Join EG and GF. ::: ED = AB (hyp.), DG AC (cons.), and <EDG = LA
(cons.), .. AEDG = ABAC (I. 4), and ... EG = BC. Again, ::: DG = DF (cons.), .-_DGF = _ DFG (1. 5), and :::_DGF is > LEGF, .. _ DFG is > LEGF, much more then is 2 EFG > LEGF, and .. EG is >EF (I. 19). But EG BC, .. BC >EF. Q.E.D.
Note.— Without the proviso marked with inverted commas, which was added by Dr Simson, the point F could occupy three positions, viz., on, above, or below EG, making three cases of the proposition. With the help of Simson's condition it can be proved that F falls below EG.
Let DF and EG meet at H. .::_DHG > L DEG (I. 16), and _ DEG is not less than
LDGE (I. 18), ...DHG> DGH, .. DG is >DH (I. 19), ... DF>DH.
.:. F falls below EG.
Assuredly the proposition is not complete without a proof of this fact.
EU. I. 26. THEOR. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either (1st) the side adjacent to the equal angles in each, or (2nd) a side opposite to one of the equal angles in each, then the two triangles shall be equal in every respect. 1st. Let triangles ABC and
F If BA be not = ED, one
of them must be greater than the other ; let BA be>
ED, cut off BG = ED (I. 3), and join GC. .:: GB = DE (cons.), BC EF (given), and LB LE (given), ..
.. AGBC = ADEF in every respect (I. 4).
= L ACB, which is impossible.
also AC = DF, and LA 2nd. Let triangles ABC and
= LF, and AB
them must be greater than the other; let BC be the greater, and cut off BH=EF (I. 3). Join AH.
:: AB = DE (given), BH = EF (cons.), and 2 B
(given),.. AABH = A DEF in every respect (I. 4). .:: LAHB = LF, but LC = _F (given), :: LAHB :
LC, which is impossible (I. 16).
.. also AC = DF, and 2 BAC = LEDF. Q.E.D. Note.- When Euclid quotes 1. 26, as in 1. 34, it is merely to prove that sides and angles of the two triangles are equal, and he makes a separate appeal to 1.4 to show that the triangles are equal (i.e., in area). This needless repetition would have been avoided if he had simply said that the triangles in 1. 20 were equal in every respect by 1. 4, as has been done above.
A similar remark applies to Euclid's use of 1. 8.
EU. I. 30. THEOR.
Straight lines which are parallel to the same straight line are parallel to one another.
Let AB be|| CD, and EF be|| CD.
-F the given lines. ::: AB is || CD, .. _ AGH = С
- HKD (1. 29). ::: EF is || CD, .. LGHF = LHKD (I. 29). .. LAGH L GHF (Ax. 1), and these are alternate angles,
.:. AB is || EF (1. 27). Q.E.D.
Exercises on Book I.-continued. 149. If a four-sided figure have all its sides equal, and one angle
a right angle, the other angles shall also be right angles (see Def. 24).
150. Parallelograms are equal in every respect which have two adjacent sides of the one equal to two adjacent sides of the other, and an angle of the one equal to a corresponding angle of the other.
151. Show that when one element (a side) is given, a square, or any regular polygon, is determined ; and that three elements (two sides and either an angle or a diagonal) determine a parallelogram.
152. How many elements are required to determine a rectangle, a rhombus, a triangle, a trapezium, and a hexagon ?
153. From three points A, B, and C in a straight line having AB = BC, perpendiculars are drawn on another straight line which does not fall between A and C; show that the sum of the perpendiculars from A and C is equal to twice the perpendicular from B. If the line falls between A and C, sum is changed to difference.
154. Of perpendiculars from the four angular points of a parallelogram upon a straight line which falls outside the parallelogram, the perpendiculars from one pair of opposite angles are together equal to those from the other pair of opposite angles.
155. If a six-sided figure have its opposite pairs of sides equal and parallel, the three diagonals shall all pass through one point.
156. In a trapezium the middle point of one of the nonparallel sides is joined with the extremities of the other, the triangle so formed is half the area of the trapezium.
157. If any point be taken in the base of a triangle and lines be drawn joining the middle points of the two parts of the base to the middle points of the sides of the triangle adjacent to the parts respectively, these two lines shall be equal.
158. If a right-angled triangle have one of the acute angles double of the other, the hypotenuse shall be double the side opposite to the smallest angle.
159. If the exterior angle and one of the interior opposite angles of one triangle be respectively double those of another, the remaining interior opposite angle of the former shall be double that of the latter.