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AF and BC, and having its base equal to AF+

BE.

AF;

For, produce BE to C, and make EC through C draw (E. 31. 1.) CD parallel to BA, and let CD meet AF produced in D; .. the figure ABCD is a ; .. (E. 34. 1.) AD BC; and (constr.) AF=EC; .. FD=BE: It is manifest, .., (from S. 54.1.) that the trapezium ABEF is the half of the ABCD; but (E. 35. 1.) the ABCD a rectangle upon the same base BC, and between the same two parallels; ... the trapezium ABEF= the half of a rectangle on the base BC, which (constr.) BE + AF, and between the two parallels BE and AF.

PROP. LVI.

73. PROBLEM. Any two parallelograms having been described on two sides of a given triangle, to apply, to the remaining side, a parallelogram, which shall be equal to their aggregate.

Let the AQ and AP be on the two sides AB, AC, of the given A ABC: It is required to apply

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equal to the AP together with the AQ. Produce QD and PE until they meet in F; join F, A; through C draw (E. 31. 1.) CG parallel to FA, and make, also, CG FA; complete the BCGH: The BCGH=□AQ+AP. For, produce FA, so that it shall meet BC in I,. and HG in K; produce, also, GC and HB, until they meet EP and DQ in L and M; .. ... the figures FACL, FABM are; and, since (constr.) the FACL, CGKI, are upon equal bases FA, CG and between the same parallels, .. (E. 36. 1.) the FACL CGKI; but (E. 35. 1.) the FACL=□ AP; .. the AP,=□GI: And in the same manner, it may be proved that the AQ=□IH; .. the whole □ BCGH BCGH-AP +□AQ.*

* If the parallelograms AP and AQ are squares, it is easy

PROP. LVII.

74. PROBLEM. A plane rectilineal figure of any number of sides being given, to find an equal rectilineal figure, which shall have the number of its sides less, or greater, by one, than that of the given figure.

First, let ABCDE be a given rectilineal figure:

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It is required to find an equal rectilineal figure, having the number of its sides less by one, than the number of the sides of ABCDE.

of the

Let A, E, D be any three consecutive given figure ABCDE; join A, D; through E draw (E. 31. 1.) EF parallel to AD and meeting CD, produced, in F; join A, F: The figure ABCF, which has the number of its sides less by one than ABCDE, is equal to ABCDE.

For, since the two AED, AEF, are upon

to shew that the parallelogram BG will also be a square; and thus the forty-seventh proposition of the first Book of Euclid's Elements will have been demonstrated.

the same base AD and (constr.) are between the same parallels, AD, EF, ... (E. 37. 1.) the ▲ AFD

AAED; to each of these equals add the figure ABCD; and the figure ABCF = the figure ABCDE.

Secondly, let ABCF be a given rectilineal figure; and let it be required to find an equal rectilineal figure, having more sides by one, than ABCF.

Take any point, D, in any of the sides, as CF, of ABCF, and join B, D, or A, D; A and B being the which are next to the F and C, at the LL extremities of CF; then, A, D having been joined, through F draw (E. 31. 1.) FE parallel to DA; and since the ADC is greater (E. 16. 1.) than the ADF, and equal (E. 29. 1.) to the ▲ EFD, .. FE falls without the given figure: In FE take any point E, and join E, A, and E, D: The figure ABCDE has more sides, by one, than the given figure ABCF; and it may be shewn, as in the preceding case, to be equal to ABCF.

75. COR. Hence, first, a triangle may be found which shall be equal to any given rectilineal figure: For the number of sides of the given figure being thus diminished, by one, at each step, they will at length be reduced to three, and the triangle which they contain, will be equal to the given figure.

Secondly, it is manifest, that, by the latter part of the preceding problem, a polygon, of any given number of sides, may be found, which shall be equal to a given triangle.

PROP. LVIII.

76. THEOREM. The diameters of any parallelogram divide it into four equal triangles.

Let ADBC be a □, of which the diameters AB,

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CD cut one another in E: The four AED, DEB, BEC, CEA are equal to one another.

For (hyp. and E. 34. 1.) the side AC of the A AEC, is equal to the side DB, of the A DEB; also (S. 42. 1.) AE= EB, and DE = EC; (E. 8. 1. and E. 4. 1.) the A AEC=A DEB. In the same manner, it may be shewn that the A AEDA CEB: And since, the two AED, AEC, stand upon equal bases DE and EC, .. (E. 38. 1.) the A AEDA AEC. It is manifest, ..., that the four AED, AEC, CEB, BED are equal to one another.

PROP. LIX.

77. PROBLEM. If two triangles have the two adjacent sides of a parallelogram for their bases, and

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