For, let PBCQ be a trapezium having the side

PQ parallel to BC, and let RD join the bisections, R and D, of the opposite sides PQ and BC: RD bisects the trapezium PBCQ.

For, join P, D, and Q, D: Then since (hyp.) PQ is parallel to BC, and that the base BD of the ▲ BPD, is equal to the base DC, of the ▲ DQC, .. (E. 38. 1.) the ▲ BPD=▲DQC; and, in the same manner, it may be shewn that the ▲ PDR = ▲ DRQ; .'., ▲ BPD + ▲ PDR = ▲ DQC+ A DRQ; i. e. the figure BPRD=CQRD; ... RD bisects the trapezium PBCQ.

PROP. LXIII.

82. PROBLEM. To bisect a given trapezium by a straight line drawn from any of its angles.

Let ABCD be a trapezium: It is required to draw a straight line from any of the , as B, which shall bisect the trapezium ABCD.

Join B, D; through A draw (E. 31. 1.) AE parallel to BD, and let CD, produced, meet AE in E; bisect (E. 10. 1.) EC in F; and join B, F; BF bisects the trapezium ABCD.

BAD,

For join B, E; and since the two BED are on the same base BD, and between the same parallels, .. (E. 37. 1.) the A BAD = A BED; to each of these equals add the ▲ BDF; .. A BAD + ▲ BDF=▲ BED +▲ BDF; i. e. the trapezium BADFA BEF; but since (constr.) EF=FC, .. (E. 38. 1.) the ▲ BEF = ABFC;.. the trapezium BADFA BFC; i.e. BF bisects the given trapezium ABCD.

PROP. LXIV.

83. PROBLEM. To bisect a given triangle by a straight line drawn through a given point in any one of its sides.

Let ABC be the given A, and let D be a given point in one of its sides BC: It is required to draw through D a straight line which shall bisect the triangle.

Bisect (E. 10. 1.) AC in E; join D, E; through

B draw (E. 31. 1.) BF parallel to DE, meeting AC in F; join D, F: DF bisects the ▲ ABC.

For join B, E and let BE cut DF in G: Then since the DFE, EBD are upon the same base

DE and (constr.) between the same parallels, .. (E. 37. 1.) the A DFE ▲ EBD; take away the common part DGE, and there remains the A BGDA EGF; to each of these equals add the trapezium ABGF, and it is manifest that the trapezium ABDF=▲ ABE; but since (constr.) AE=EC, .. (E. 38. 1.) the ▲ ABE= ▲ EBC; ... the trapezium ABDF is equal to the half of the given ▲ ABC; i. e. DF bisects the ▲ ABC.

PROP. LXV.

84. PROBLEM. Equal triangles, which have their bases in the same straight line and which are

between the same parallels, stand upon equal

bases.

For if not, let one of the bases be greater than the other; .. (S. 60. 1.) the ▲, of which it is the base, is greater than the other, which is contrary to the supposition: .., neither of the bases can be greater than the other; i. e. the bases are equal to one another.

PROP. LXVI.

85. PROBLEM. To describe a parallelogram, the surface and perimeter of which shall be respectively equal to the surface and perimeter of a given triangle.

Let ABC be the given A: It is required to

describe a, which shall be equal to the ▲ ABC, and which shall, also, have its perimeter equal to the perimeter of ABC.

Bisect (E. 10. 1.) BC in D; produce AC to E, and make CEAB; bisect AE in F; through A draw (E. 31. 1.) AH parallel to BC; from Das a centre, at a distance equal to AF describe a circle, cutting AH in G ; join D, G, and through C draw CH parallel to DG: Then is the DCHG=A ABC, and the perimeter of DCHG is, also, equal to the perimeter of ABC.

For join A, D; and since BD = DC, ... (E: 38. 1.) the ▲ ABD =▲ ACD, so that the whole A ABC is the double of the AADC: Again, since the DCHG and the ▲ ADC are on the same base DC, and between the same parallels, .`. (E. 41. 1.) the DCHG is the double of the A ADC; as is, also, the ▲ ABC: .. the DCHGA ABC: And because (constr.) DG is equal to the half of BA+AC, and that (E. 34. 1.) CH=DG, ... DG +CH=BA + AC; also (E. 34. 1.) GH=DC= DB; .. DC + GH=BD + DC = BC; .. DG + GH+HC+CD=BA+AC+CB.

86. THEOREM.

PROP. LXVII.

The two triangles formed by drawing straight lines, from any point within a parallelogram, to the extremities of either pair of opposite sides, are, together, half of the parallelogram.