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 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,

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 ▲ DEB; also (S. 42. 1.) AE EB, and DE EC; (E. 8. 1. and E. 4. 1.) the ▲ AEC=A DEB. In the same manner, it may be shewn that the ▲ AED= A 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 adja cent sides of a parallelogram for their bases, and

have their common vertex situated in the diameter, or in the diameter produced, they shall be equal to one another.

Let the two AFC, BFC, have the two adjacent sides AC, BC, of the

ADBC, for their

F

E

B

C

bases, and also have their common vertex situated at any point F, in the diameter DC, or in DC, produced: The ▲ AFCA BFC.

First, let the point F be in the diameter DC: Join A, B; and let AB cut DC in E.

Then, since (S. 42. 1.) AE=EB, .. (E. 38. 1.) the A AECA BEC, and the ▲ AEFA BEF; `.. the ▲ AFC, BFC, which are the differences of these equals, are equal to one another.

And the proposition may, in the same manner, be shewn to be true, when the common vertex of the two, which have AC and BC for their bases, is in DC produced.

PROP. LX.

78. THEOREM. Of all triangles, which are between the same parallels, that which stands on the greatest base is the greatest.

For it is manifest, that the A which has the greater base will exceed the A which is formed by joining its vertex and the extremity of a segment of its base made equal to the base of the other ▲ But the ▲ so formed is equal (E. 38. 1.) to the other given A; .. the A which has the greater base is greater than that other triangle.

PROP. LXI.

79. THEOREM. The straight line, joining the vertex and the bisection of the base of any triangle, bisects every other straight line that is parallel to the base and is terminated by the two remaining sides of the triangle.

Let PQ be any straight line, either within or

without the ▲ ABC, parallel to the base BC, and let AD, joining the vertex A and the bisection D of BC, cut PQ in R: PQ is bisected by AD in R.

First, let PQ be within the ▲ ABC; and if PR be not equal to RQ, one of them is the greater: Let PR > RQ; and join D, P, and D, Q.

Then since (hyp.) the base BD, of the ▲ BAD, is equal to the base DC, of the A CAD, .. (E. 38. 1.) the ▲ BAD A CAD; also, because BD=DC, and that (hyp.) PQ is parallel to BC, .. (E. 38. 1.) the ▲ BPD ACQD; if, ..., the two latter equal be taken from the equal ▲ BAD, CAD, there remains the ▲ APD = A AQD: But, since PR > RQ, the ▲ APR > A AQR, and the ▲ DPR > ▲ DQR; ..., the whole ▲ ·APD> AAQD; but it has been shewn that the AAPD AAQD; and it is, also, greater; which is absurd: .., neither of the two lines PR, RQ, can be greater than the other; .., PR=RQ. In a similar manner the proposition may be proved, when PQ is without the ▲ ABC.

80. COR. Hence, it is easily shewn, ex absurdo, that the straight line joining the bisections of any two straight lines, that are parallel to the base, and terminated by the sides of a A, passes through the vertex of the A.

PROP. LXII.

81. THEOREM. If two opposite sides of a trapezium be parallel to one another, the straight line, joining their bisections, bisects the trapezium.