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PROPOSITION XL. THEOREM.
Equal triangles, on equal bases in the same straight line, and on the same side of it, are between the same parallels.
Let the = As ABC, DEF be on = bases BC, EF, in the same straight line BF, and on the same side of it.
Then they shall be between the same ||s.
Constr. Join AD; AD shall be || BF.
meeting ED at G, and join GF. Dem. Then A ABC= AGEF, :;: they are
on = bases BC, EF, and between the same ||s BF, AG; (i. 38) but a ABC = A DEF; (hyp.).. also A DEF = A GEF,
(ax. 1) the greater = the less, which is impossible.
.. AG is not || BF. And in the same manner it can be demonstrated that no other st. line through A but AD is || BF;
.. AD is || BF. Therefore, equal triangles on, &c.
Q. E. D.
EXAMPLES ON PROPOSITIONS 39 AND 40.
1. A quadrilateral is divided by its diagonals into four triangles of which two opposite ones are equal. Show that two sides of the quadrilateral are parallel.
2. The straight line which joins the middle points of two sides of a triangle is parallel to the base.
3. The straight line which joins the middle points of two sides of a triangle is equal to half the base.
4. Given the middle points of the sides of a triangle, to construct it.
5. If the points of bisection of the sides of a triangle be joined, the triangle so formed will be one fourth of the original triangle.
6. Two straight lines, drawn to bisect the opposite sides of any quadrilateral, will also bisect one another.
PROPOSITION XLI. THEOREM.
If a parallelogram and a triangle be on the same base, and between the same parallels, the parallelogram is double of the triangle.
Let the O ABCD and the A EBC be on the same base BC and between the same || BC, AE.
Then shall ABCD be double of A EBD.
B Constr. Join AC. Dem. Then A ABC = A EBC, ::: they are on the same
base BC, and between the same || BC, AE. (i. 37) But ABCD is double of A ABC, : AC bisects the ;
(i. 34) .:. ABCD is also double of EBC. Therefore, if a parallelogram, &c. Q. E. D.
1. O is any point within a parallelogram ABCD, and lines are drawn from it to the angles. Show that the triangles OAB, OCD are together equal to the triangle ABC.
2. If a parallelogram is double of a triangle, and they have the same base, or equal bases on the same straight line, and towards the same parts, they shall be between the same parallels.
PROPOSITION XLII. PROBLEM.
To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given angle.
Let ABC be the given A and D the given L.
It is required to describe a = the given A, having one of its Zs= D.
Constr. Bisect BC in E, (i. 10) and join AE;
at E make < CEF = LD; (i. 23) through A draw AFG || BC, and through C draw CG || EF;
(i. 31) Dem. Then CEFG is a O.
(def.) And :: As ABE, AEC are on= bases BE, CE, and between the same lis BC, AG, they are equal;
(i. 38) ..A ABC is double of A AEC. But O FECG is double of A AEC, : they are on the same base and between the same lls ; (i. 41)
..OFECG = A ABC, (ax. 6) and has one of its Zs CEF = the given _ D.
(constr.) Therefore, a parallelogram has been described as required.
Q. E. F. Describe a parallelogram equal to a given equilateral triangle having an angle equal to an angle of the triangle. Show that the perimeters are likewise equal.
PROPOSITION XLIII, THEOREM.
The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another.
Let ABCD be a O, of which AC is a diameter, and EH, GF the Os about AC (that is, through which AC passes); and BK, KD the others which make up the whole figure, and which are therefore called the complements.
Then shall complement BK = complement KD.
Dem. .:: ABCD is a O and AC its diameter, ., A ABC = A ADC.
(i. 34) Again, :: EKHA is a O and AK its diameter, .. A AEK = S AHK;
(i: 34) and for the same reason 3 KGC = A KFC; i. the two As AEK, KGC = the two As AHK, KFC;
(ax. 2) but the whole A ABC = the whole » ADC; .. the remainder, the complement BK = the remainder, the complement KD.
(ax. 3:) Therefore, the complements, &c.
Q. E. D.