For since FH is parallel to AD,

FH AD :: HE: DE,

:

but AD: DE :: DE: (DC=) HG,

.. FH DE: HE: HG.

(41.) If from the extremities of the base of any triangle, two straight lines be drawn intersecting each other in the perpendicular, and terminating in the opposite sides; straight lines drawn from thence to the intersection of the perpendicular with the base, will make equal angles with the base.

From A and C, the extremities of AC, the base of the triangle ABC, let AE, CF be drawn intersecting the perpendicular BD in the same point G. Join FD, ED; these lines make equal angles FDA, EDC with the base.

H

M

B

G

Draw EI, FH perpendicular, and KGL parallel to

the base, then FH is parallel to BD,

and . BG: BD :: FM: FH.

And in the same manner it may be shewn that

BG BD::EN: EI;

whence FM: FH:: EN

EI;

and .. FM: EN :: FH : EI.

But FM: EN :: FG: GN :: KG: GL :: HD: DI,

:. HD : DI :: HF : EI,

whence the two triangles DFH, DEI, having the angle at H equal to the angle at 1, and the sides about the equal angles proportional, are equiangular; .. the angle HDF is equal to EDI.

(42.) In every triangle, the intersection of the perpendiculars drawn from the angles to the opposite sides, the intersection of the lines from the angles to the middle of the opposite sides, and the intersection of the perpendiculars from the middle of the sides, are all in the same straight line. And the distances of those points from one another are in a given ratio.

From the angles A and B of the triangle ABC, let AD, BE be drawn perpendicular to the opposite sides, H will be the intersection of the three perpendiculars (vii. 34.). From A and B

H

G

draw AG, BF to the points of bisection of the opposite sides, intersecting in K, which.. is (iii. 28.) the intersection of the lines drawn from the angles to the middle of the opposite sides; and from F and G draw the perpendiculars FI, GI meeting in I, which.. (iii. 27.) is the intersection of the three perpendiculars. Join HK, KI; HKI is a straight line.

Join GF. (Eucl. vi. 2.) AB is parallel to, and double of GF;.. by similar triangles ABK, KFG, BK is double of KF, and AK double of KG. And the triangles AHB, FIG are equiangular, .. AH is double of IG, and BH is double of IF;

and .. BH: IF :: 2 : 1 :: BK : KF, whence the triangles BHK, KIF having the angles at B and F equal, and the sides about them proportional, are similar, .. the angle HKB is equal to IKF, .. H, K, and I are in the same straight line.

And since BK is double of KF, HK is double of KI, and... their distances from each other will be in an invariable ratio.

(43.) If straight lines be drawn from the angles of a triangle through any point, either within or without the triangle, to meet the sides, and the lines joining these points of intersection and the sides of the triangle be produced to meet; the three points of concourse will be in the same straight line.

Let ABC be a triangle from the three angles of which let lines AF, BE, CD be drawn through a point P within the triangle. Join DE, DF, EF, and produce them

H

2

to meet the sides in H, G, I; these three points will be in the same straight line.

Join GH, HI. Then the three angles of the triangle DHG being equal to two right angles, as also the three EHI, EIH, and (HEI or) DEF, as also the two DFI, GFI; .. the three angles of the triangle DHG together with the angles EHI, EIH, Def, DFI, GFI are equal to six right angles. Now the angles of the triangles DEF, FGI are together equal to four right angles, whence DHG, DHI are equal to two right angles; or GH, HI are in the same straight line.

SECT. IV.

(1.) The diameters of a rhombus bisect each other at right angles.

Let ABCD be a rhombus, whose diameters are AC, BD; they bisect each other at right angles in E.

B

A

E

D

Since AB-AD, and AC is common to the two triangles ABC, ADC, the two BA, AC are equal to the two DA, AC, each to each, and BC= DC, .. the angle BAC is equal to the angle DAC. Again, since BA, AE are equal to DA, AE, each to each, and the included angles are equal,.. BE=ED, and the angles AEB, AED are equal, and .. are right angles. For the same reason AE=EC; also the angles BEC, DEC are right angles.

(2.) If the opposite sides or opposite angles of a quadrilateral figure be equal, the figure will be a parallelogram.

Let ABCD be a quadrilateral figure, whose opposite sides are equal. Join BD. Since AB DC, and BD is common, the two AB, BD are equal to

B

the two CD, DB, each to each, and AD= BC, .. the angle ABD = BDC, whence (Eucl. i. 27.) AB is parallel to DC; also the angle ADB = DBC, whence AD is parallel to BC; and the figure is a parallelogram.

Again, let the opposite angles be equal. Then since the four angles of the quadrilateral figure ABCD are equal to four right angles, and that BAD, ADC together are equal to DCB, CBA, .. BAD, ADC together are equal to two right angles; whence AB is parallel to CD. In the same way it may be shewn that AD is parallel to BC, and.. ABCD is a parallelogram.

(3.) To bisect a parallelogram by a line drawn from a point in one of its sides.

Let ABCD be a parallelogram, and P a given point in the side AB. Draw the diameter BD, which bisects the parallelogram. Bisect BD

E

F

in F; join PF, and produce it to E. PE bisects the parallelogram.

Since the angle PBD is equal to the angle BDE, and the vertically opposite angles at Fare equal, and BF = FD, . the triangles PBF, DFE are equal. But the triangle ABD is equal to BDC, .. APFD is equal to BFEC; and to these equals adding the equal triangles DFE, PFB, the figure APED=PECB; and AC is.. bisected by PE.

COR. Any line drawn through the middle point of the diameter of a parallelogram is bisected in that point.

(4.) If from any point in the diameter (or diameter produced) of a parallelogram straight lines be drawn to the opposite angles; they will cut off equal triangles.

Q