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,

A

H

Since AH= EB, and AE=BF, and Ꭰ G the angles at A and B are 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.

B

E

P

C

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.

For (Eucl. vi. 1.) AEB BEC: AE

:

EC,

and AED: DEC :: AE : EC,

.. AEB BEC:: AED: DEC.

(13.) If two opposite angles of a trapezium be right ungles; the angles subtended by either side at the two opposite angular points will be equal.

Let the two angles ACB, ADB of the trapezium ACBD, be right angles. Join AB, CD; the angles ACD, ABD, subtended by AD, are equal.

B

Bisect AB in E. Join CE, ED, and produce CE to F. Then (iii. 2.) AE, EB, EC, ED are equal to one another. Also the angle AED is equal to the two EDB, EBD, i. e. to twice EBD; and DEF is equal to the two DCE, EDC, i. e. to twice DCE; and AEF is equal to twice ACE; .. twice ACE and twice ECD, or twice ACD will be equal to AED, i, e. to twice EBD, .. ACD = ABD.

The same may be proved for the angles standing on any of the other sides.

(14.) To determine the figure formed by joining the points of bisection of the sides of a trapezium; and its ratio to the trapezium.

Let ABCD be a trapezium, whose sides are bisected in E, F, G, H. Let the points of bisection be joined ; and draw the diagonals AC, BD.

Since AB, AD are bisected in E and H, (Eucl. vi. 2.)

EH is parallel to BD; and for the same reason FG is parallel to BD, and .. to EH. In the same way it may be shewn that EF is parallel to HG, and .. the figure EFGH is a parallelogram.

Again (Eucl. vi. 19.) the triangle EBF is to the triangle ABC in the duplicate ratio of EB : AB, i. e. in the ratio of 1: 4, .. EBF is equal to one fourth of ABC; for the same reason HDG is one fourth of DAC, whence EBF and HDG are together equal to one fourth of the trapezium. For the same reason HAE and GFC are together equal to one fourth of the trapezium; therefore the four triangles together are equal to half the trapezium; and consequently HEFG is equal to half of ABCD.

COR. 1. Hence two lines, drawn to bisect the opposite sides of a trapezium, will also bisect each other.

COR. 2. If the sides of a square be bisected and the points of bisection joined, the inscribed figure is a square, and equal to half the original square.

(15.) To determine the figure formed by joining the points where the diagonals of the trapezium cut the parallelogram; and its ratio to the trapezium.

Let I, K, L, M be the points of intersection; (see last Fig.) join IK, KL, LM, MI. And let O be

the intersection of the diagonals. Since EK is parallel to OI,

BK : KO :: BE : EA, i. e. in a ratio of equality. For the same reason AI-10. Whence the sides of the triangle AOB being cut proportionally, IK is parallel to AB. In the same manner it may be shewn that KL, LM, MI are respectively parallel to BC, CD, DA; whence the figure IKLM will be similar to ABCD, Also since the triangle MIK is half the parallelogram HK, and MLK half of GK, .. the figure IKLM is half of HF, and .. equal to one fourth of the trapezium ABCD.

(16.) If two sides of a trapezium be parallel; its area is equal to half that of a parallelogram, whose base is the sum of those two sides, and altitude the dicular distance between them.

Let ABCD be a trapezium, whose side AB is parallel to DC. Produce DC to E, making CE=AB. Draw BF, and CG, EH parallel to AD,

perpen

meeting AB produced. Then AE is a parallelogram of the same altitude with the trapezium, and its base is equal to the sum of the sides AB, DC; and ABCD is half of AE.

Since DF=CE, the parallelograms AF, GE are equal (Eucl. i. 36.); and the diameter BC bisects the parallelogram FG; whence ABCD=BCEH, and .. the trapezium is half of the parallelogram AE.

R