of ABCD. In the same manner if a line be drawn through P parallel to AB or DC, it may be shewn that APB, DPC together are half of ABCD.

(8.) If a straight line be drawn parallel to one of the sides of a parallelogram, and one extremity of this line be joined to the opposite one of the parallel side, by a line which also cuts the diameter; the segments of the diameter made by this line will be reciprocally proportional to the segments of that part of it which is intercepted between the side and the parallel line.

Let EF be drawn parallel to AD one of the sides of the parallelogram ABCD, cutting the diameter BD in G. Join AF, cutting it also in H; then will BH: HD :: HD: HG.

H G

E

B

For the angle ABH being equal to HDF, and AHB to DHF, the triangles AHB, DHF are equiangular, and ..BH: HD :: AH : HF :: DH: HG, since the triangles AHD, FHG are also equiangular.

(9.) If two lines be drawn parallel and equal to the adjacent sides of a parallelogram; the lines joining their extremities, if produced, will meet the diameter in the same point.

Let HI, FG be drawn equal and parallel to the adjacent sides AB, BC of the parallelogram ABCD. Join HF, GI; these lines produced will meet the diameter DB in the same point.

Produce AB, CB to K and L. Then the triangles AFH, LBF having the vertically opposite angles at F equal, and the alternate angles AHF, FLB also equal, are equiangular,

whence AF FB :: (HA=) IB: BL; and in the same manner it may be shewn that (GC=) FB: CI :: BK: BI,

.. AF: CI :: BK : BL.

But AF= DG, and CI=DH, .. DG : DH :: BK : BL, and .. HF, DB, GI converge to the same point.

(10.) If in the sides of a square, at equal distances from the four angles, four other points be taken, one in each side; the figure contained by the straight lines which join them shall also be a square.

Let E, F, G, H be four points at equal distances from the angles of the square ABCD. Join EF, FG, GH, HE; EFGH is also a square.

H

Since AH-EB, and AE=BF, and D G the angles at A and B are at right angles, .'. HE= EF, and the angle AEH is equal to the angle BFE. In the same way it may be shewn that HG and GF are

of ABCD. In the same manner if a line be drawn through P parallel to AB or DC, it may be shewn that APB, DPC together are half of ABCD.

(8.) If a straight line be drawn parallel to one of the sides of a parallelogram, and one extremity of this line be joined to the opposite one of the parallel side, by a line which also cuts the diameter; the segments the diameter made by this line will be reciprocally proportional to the segments of that part of it which is intercepted between the side and the parallel line.

Let EF be drawn parallel to AD one of the sides of the parallelogram ABCD, cutting the diameter BD in G. Join AF, cutting it also in H; then will BH: HD :: HD: HG.

H

E

B

For the angle ABH being equal to HDF, and AHB to DHF, the triangles AHB, DHF are equiangular, and ..BH : HD :: AH : HF :: DH: HG, since the triangles AHD, FHG are also equiangular.

(9.) If two lines be drawn parallel and equal to the adjacent sides of a parallelogram; the lines joining their extremities, if produced, will meet the diameter in the same point.

Let HI, FG be drawn equal and parallel to the adjacent sides AB, BC of the parallelogram ABCD. Join HF, GI; these lines produced will meet the diameter DB in the same point.

Produce AB, CB to K and L. Then the triangles AFH, LBF having the vertically opposite angles at F equal, and the alternate angles AHF, FLB also equal, are equiangular,

whence AF: FB :: (HA=) IB : BL; and in the same manner it may be shewn that (GC=) FB : ÇI :: BK : BI,

.. AF : CI :: BK : BL.

But AF=DG, and CI=DH, .. DG : DH :: BK : BL, and .. HF, DB, GI converge to the same point.

(10.) If in the sides of a square, at equal distances from the four angles, four other points be taken, one in each side; the figure contained by the straight lines which join them shall also be a square.

Let E, F, G, H be four points at equal distances from the angles of the square ABCD. Join EF, FG, GH, HE; EFGH is also a square.

H

D G

C

Since AH EB, and AE = BF, and the angles at A and B are at right angles, .'. HE= EF, and the angle AEH is equal to the angle BFE. In the same way it may be shewn that HG and GF are

each of them equal to HE and EF, .. the figure HEFG is equilateral. It is also rectangular; for since the exterior angle FEA is equal to the interior angles EBF, EFB; parts of which AEH and EFB are equal; .. the remaining angle FEH is equal to the remaining angle FBE, and .. is a right angle. In the same manner it may be shewn that the angles at F, G, H are right angles, and .. EFGH being equilateral and rectangular, is a square.

(11.) The sum of the diagonals of a trapezium is less than the sum of any four lines which can be drawn to the four angles from any point within the figure, except from the intersection of the diagonals.

Let ABCD be a trapezium, whose diagonals are AC, BD, cutting each other in E; they are less than the sum of any four lines which can be drawn to the angles from any other point within the trapezium.

Take any point P, and join PA, PB, PC, PD. Then (Eucl. i. 20.) AC is less than AP, PC; and BD is less than BP, PD; .. AC, BD are less than AP, PB, PC, PD.

(12.) Every trapezium is divided by its diagonals into four triangles proportional to each other.

Let ABCD be a trapezium (see last Fig.) divided by its diagonals AC, BD into the triangles AEB, BEC, AED, DEC; these are proportional to each other.