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.

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.

Let AC be greater than AB; A

B

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

is common and at right 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

A

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.

A

F

E

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

B

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.

(6.) If from one of the equal angles of an isosceles triangle any line be drawn to the opposite side, and from the same point, a line be drawn to the opposite side produced, so that the part intercepted between them may be equal to the former; the angle contained by the side of the triangle and the first drawn line is double of the angle contained by the base and the latter.

Let ABC be an isosceles triangle, having the side AB equal to AC. From B draw any line BD, and also BE cutting off DE equal to DB; the angle ABD is double of CBE.

B

A

D

C

E

For the angle DCB is equal to the two DEB, CBE, i. e. to the two DBE, CBE, or to DBC and twice CBE; but DCB is equal to ABC, .. ABC is equal to DBC and twice CBE, and taking away the angle DBC, which is common to both, the angle ABD is equal to twice CBE.

(7.) If from the extremity of the base of an isosceles triangle, a line equal to one of the sides be drawn to meet the opposite side; the angle formed by this line and the base produced, is equal to three times either of the equal angles of the triangle.

Let ABC be an isosceles triangle having the side AB equal to AC. From C to AB (produced if necessary) draw CD equal to AC, and let BC be produced; the angle DCE is equal to three times the angle ABC.

Since CA is equal to CD, the angle CAD is equal to CDA, .. CDA and twice ABC are together equal

to two right angles, and .. are equal to CDA, CDB; whence CDB is double of ABC. Now (Eucl. i. 32.) the angle DCE is equal to the two angles CDB, CBD and consequently is equal to three times the angle ABC.

(8.) The sum of the sides of an isosceles triangle is less than the sum of the sides of any other triangle on the same base and between the same parallels.

Let ACB be an isosceles triangle, and ADB any other triangle on the same base, and between the same pa

E

rallels AB, ED; AC and CB together will be less than AD and DB.

A

B

Since EC is parallel to AB, the angle ECA is equal to CAB; and for the same reason DCB is equal to CBA; but CAB being equal to CBA, ECA is equal to DCB; .. AC and BC drawn from two given points A and B on the same side of the line ECD given in position make equal angles with the line, .. (i. 6.) they are together less than any other two lines AD, DB, drawn from the same points to that line.

(9.) If from one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side; the part of it intercepted by a perpendicular from the vertex will have to one of the equal sides, the same ratio that the segment of the base has to the perpendicular upon the base.