From any point E, in AC

B

C

the diameter of the parallelo

gram ABCD, let lines EB, A

ED be drawn; the triangles ABE, AED are equal; as also the triangles BEC, CED.

Draw the diameter BD. The bases BF, FD being equal, the triangles BFA, DFA (Eucl. i. 38.), as also the triangles BFE, DFE are equal, hence .. BAE, DAE are equal. And ABC being equal to ADC, the triangles BEC, DEC are also equal.

(5.) From one of the angles of a parallelogram to draw a line to the opposite side, which shall be equal to that side together with the segment of it which is intercepted between the line and the opposite angle.

Let ABCD be the parallelogram, A the angle from which the line is to be drawn. Produce DC to E, making CE CD. Join AE, and at the point A make the angle EAF AEF; AF is the line required.

=

=

B

F

For CE being equal to CD, EF = DC and CF together; and the angles FEA, FAE being equal, FA= FE, and .. AF=DC and CF together.

COR. In the same manner if CE=CB, AF=EF= BC and CF together.

(6.) If from one of the angles of a parallelogram a straight line be drawn cutting the diameter and a side

produced; the segment intercepted between the angle and the diameter, is a mean proportional between the segments intercepted between the diameter and the sides.

From B, one of the angles of the parallelogram ABCD, let any line BE be drawn cutting the diameter AG in F, the opposite side in G, and

A

B

H

F

G

E

AD produced in E; BF is a mean proportional between FG and FE.

Draw DH parallel to BF, and .. equal to it; .. also AH-FC, and AF CH. Since HD is parallel to BG, FG DH (:: CF: CH :: AH : AF) :: DH : FE, or FG BF :: BF: FE.

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

Let P be any point within the parallelogram ABCD, from which let lines PA, PD, PB, PC be drawn to the extremities of the oppo

site sides; the triangles PAD,

A

F

E

B

PBC are equal to half the parallelogram; as also the triangles APB, DPC.

Through E draw EPF parallel to AD or BC; then (Eucl. i. 41.) the triangle APD is half of AEFD, and BPC is half of BEFC, .. APD, BPC are together half

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.

D

H G

B

For the angle ABH being equal to HDF, and AHB 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,

H

K

D

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 AHEB, and AE=BF, and the angles at A and B are right angles, . HEEF, 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.

A

B

E

P

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.