together equal to BKC, CQP, i. e. (Eucl. i. 29.) to two right angles, whence (Eucl. i. 14.) PQ and QR are in the same straight line.

(99.) If from the extremities of the diameter of a circle any number of chords be drawn, two and two intersecting each other in a perpendicular to that diameter; the lines joining the extremities of every corresponding two will meet the diameter produced in the same point.

From A and B, the extremities of the diameter AB of a semicircle, let AC, BD be drawn intersecting each

other in FH, which is perpendicular to AB. Join CD, and produce it to meet BA in P; P is a fixed point, or the line joining the extremities of every other two chords intersecting each other in FH will pass through P.

Join BC; and bisect BG in O; and with the centre O, and radius OB, describe a circle HGB, which will circumscribe the quadrilateral figure HGCB. Take E the centre of the semicircle, and join HC, EC. The angle PCE is equal to PCA, ACE together, i. e. to DBA, CAE together; and the angle CHE is equal to ACH, CAH together, i.e. to DBA, CAH together, ... PCE = CHE, and the angle at E being common, the triangles CEH, CPE are equiangular;

whence EH: EC :: EC: EP, in which proportion the three first terms being invariable, EP is also, and the point E being fixed, P is also.

(100.) If from a given point in the diameter of a semicircle produced, three straight lines be drawn, one of which is inclined at a given angle to the diameter, another touches the semicircle, and the third cuts it, in such a manner, that the distance of the given point from the nearer extremity of the diameter, and the perpendiculars drawn from that extremity on the three aforesaid lines may be proportional; then will the lines, which join the extremities of the diameter and of that part of the cutting line which is within the circle, intersect each other in an angle equal to the given angle.

From a given point C, in the diameter AB produced of the semicircle AGB, draw CD inclined at a given

H

K

E

I F

angle to AC, CG touching, and CIH cutting the circle in such a manner that BD, BE, BF being drawn respectively perpendicular to them, CB may be to BD as BE to BF; then if AI, BH be joined, the angle ALH or BLI will be equal to BCD.

Join OH, OG; and draw OK perpendicular to HI. Now the angles at E and F being right angles, as also those at G and K, BE is parallel to OG, and BF to OK; .. (OG) OH: BE: CO: CB:: OK: BF,

.. OH OK :: BE : BF :: BC : BD; also the angle at D is equal to OKH, .. (Eucl. vi. 7.) the triangles OHK, BCD are equiangular, and the angle OHK is equal to BCD. But OHB is equal to OBH,

i. e. to AIH (Eucl. iii. 21.), .. OHK is equal to AIH, LHI together, i. e. to ALH (Eucl. i. 32.); wherefore ALH is equal to BCD.

The Straight line, and trigny ks

SECT. III.

(1.) Any side of a triangle is greater than the difference between the other two sides.

Let ABC be a triangle; any of

its sides is greater than the difference of the other two.

B

Let AC be greater than AB; and cut off AD=AB; join BD; then the angle ABD is equal to ADB. But the exterior angle BDC is greater than DBA, i. e. than BDA, and .. greater than DBC (Eucl. i. 16.); whence BC is greater than DC, i. e. than the difference of the sides AC and AB. In the same way it may be shewn that AB is greater than the difference of AC and BC; and AC greater than the difference of AB and BC.

(2.) In any right-angled triangle, the straight line joining the right angle and the bisection of the hypothenuse is equal to half the hypothenuse.

Let ACB be a right-angled triangle, whose hypo

thenuse AB is bisected in D; join DC; DC is equal to the half of AB.

From D draw DE parallel to AC, .. (Eucl. vi. 2.) BE=EC, and ED B

is common and at right angles to BC, angles to BC, .. DC≈BD, i. e. the half of AB.

(3.) If from any point within an equilateral triangle perpendiculars be drawn to the sides; they are together equal to a perpendicular drawn from any of the angles to the opposite side.

From any point D within the equilateral triangle ABC let perpendiculars DE, DF, DG be drawn to the sides, they are together equal to BH a perpendicular drawn from B on the opposite side AC.

B

F

HE C

Join DA, DB, DC. Since triangles upon the same and equal bases are to one another as their altitudes, ABC ADC :: BH : DE,

:

also ABC BDC :: BH: DF

and ABC: ADB :: BH: DG;

whence ABC : ADC+BDC+ADB :: BH : DE + DF+ DG, in which proportion the first term being equal to the second, .. DE+DF+DG=BH.

(4.) If the points of bisection of the sides of a given triangle be joined; the triangle so formed will be one fourth of the given triangle.

Let the sides of the triangle ABC be bisected in the points D, E, F; join DE, EF, FD; the triangle DEF is one fourth of the triangle ABC.

Since AB and AC are bisected in D and F, (Eucl. vi. 2.) DF is parallel to BC; and for the same reason FE is parallel to AB, and DFEB is a parallelogram, .. the triangle DFE is equal to DBE. In the same way it may be shewn to be equal to FEC and ADF; and.. it is one fourth of ABC.

(5.) The difference of the angles at the base of any triangle is double the angle contained by a line drawn from the vertex perpendicular to the base, and another bisecting the angle at the vertex.

From B the vertex of the triangle ABC let BE be drawn perpendicular to the base, and BD bisecting the angle ABC; the difference of the angles BAC, BCA is double the angle EBD.

A

ED

The angle BAC is equal (Eucl. i. 32.) to the difference of the angles BEC and ABE, i. e. of a right angle and ABE. Also the angle BCA is equal to the difference of a right angle and EBC, .. the difference of the angles BAC and BCA is equal to the difference of the angles ABE and EBC, i. e. (since ABD=DBC) to twice the angle EBD.