THEOREM IX.

If from the centre of a circle ABE there be drawn a perpendicular CD on the chord AB, and produced till it meets the circle in F, that line CF, will bisect the arc AB in the point F. fig. 29.

Let the lines AF and BF be drawn, then in the triangles ADF, BDF: AD=BD (by the last;) DF is common, and the angle ADF-BDF being both right, for CD or DF is a perpendicular. Therefore (by theo. 6.) AF FB; but in the same circle, equal lines are chords of equal arcs, since they measure them (by def. 19.) whence the arc AF=FB, and so AFB is bisected in F, by the line CF.

Cor. Hence the sine of an arc is half the chord of twice that arc. For AD is the sine of the arc AF and AF is half the arc, and AD half the chord AB (by theo. 8.) therefore the cor. is plain.

THEOREM X.

fig. 30.

In any triangle ABD, the half of each side is the sine of the oppo

site angle. fig. 30.

C

Let the circle ADB be drawn thro' the points A, B,

BKD (by cor. 1. theo. 7.) viz. the chord of BK is the measure of the angle BAD; therefore (by cor. to the last) BE the half of BD is the sine of BAD; the same way may be proved, that half of AD is the sine of ABD, and the half of AB the sine of ADB. Q. E, D.

THEOREM XI.

fig. 22,

I

E

C

If a right line GH cut two other A right lines AB, CD, so as to make the alternate angles AEF, EFD equal to each other, then the lines AB and CD will be parallel. fig. 22.

-D

If it be denied that AB is parallel to CD, let IK be parallel to it; then IEF=(FED)=AEF (by part 2. theo. 3.) a greater to a less, which is absurd, whence IK is not parallel; and the like we can prove of all other lines but AB; therefore AB is parallel to CD. Q. E. D.

THEOREM XII.

If two equal and parallel lines AB, CD be joined by two other lines AD, BC, those shall be also equal and parallel. fig. 3.

Let the diameter or diagonal BD, be drawn and we will have the two triangles ABD, CBC, whereof AB in one is = to CD in the other, DB common to both, and the angle ABD=CDB (by part 2. theo. 3.) therefore (by theo. 6.) AD=CB, and the angle CBD=

ADB, and thence the lines AD and BC are parallel by the preceding theorem.

Cor. 1. Hence the quadrilateral figure ABCD is a parallelogram, and the diagonal BD bisects the same, in as much as the triangle ABD BDC, as now proved.

Cor. 2. Hence also the triangle ADB on the same base AB, and between the same parallels with the pa rallelogram ABCD, is half the parallelogram.

Cor. 3. It is hence also plain, that the opposite sides of a parallelogram are equal; for it has been proved that ABCD being a parallelogram, AB will be-CD and AD-BC.

THEOREM XIII.

All parallelograms on the same or equal bases and between the same parallels, are equal to one another, that is, if BD=GH, and the lines BH and AF parallel, then the parallelogram ABDC=BDFE=EFHG. fig. 31.

For AC=DB+EF (by cor. the last ;) to both add CE, then AE-CF. In the triangles ABE, CDF: AB-CD and AE-CF and the angle BAE=DCF (by part 3. theo. 3.) therefore the triangle ABE= CDF (by theo. 6.) let the triangle CKE be taken from both, and we will have the trapezeum ABKC=KD

the parallelogram ABCD=BDEF; in like manner we may prove the parallelogram EFGH=BDEF. Wherefore ABCD=BDEF=EFHG. Q. E. D.

Cor. Hence it is plain that triangles on the same or equal bases and between the same parallels, are equal, seeing (by cor. 2. theo. 12.) they are the halves of their respective parallelograms.

Through A draw AKL perpendicular to the hypothenuse BC, join AH, AM, DC and BG; in the triangles BDC, ABH, BD=BA being sides of the same square, and also BC=BH, and the included angle DBC=ABH, (for DBA=CBH being both right, to both add ABC, then DBC=ABH) therefore the triangle DBC=ABH (by theo. 6.) but the triangle DBC is half of the square ABDE (by cor. 2. theo. 12.) and the triangle ABH is half the parallelogram BKLH (by the same;) therefore half the square ABDE is equal to half the parallelogram BKLH, and the square ABDE equal to the parallelogram BKLH.

=

The same way it may be proved, that the square ACGF, is equal to the parallelogram KCLM. So ABDE+ACGF the sum of the squares, BKLH+ KCML, the sum of the two parallelograms or square BCMH; therefore the sum of the squares on AB and AC is equal to the square on BC. Q. E. D.

Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having the legs; thus, the square root of the sum of the squares of the base and perpendicular, will be the hypothenuse.

Cor. 2. Having the hypothenuse and one leg given to find the other; the square root of the difference of the squares of the hypothenuse and given leg, will be the required leg.

THEOREM XV.

In all circles the chord of 60 degrees is always equal in length to the radius.

In the triangle ABC, the angle ACB is 60 degrees, being measured by the arc AEB; therefore the sum of the other two angles is 120 degrees (by cor. theo. 5.) but since AC=CB, the angle CAB=CBA (by lemma preceding theo. 7.) consequently each of them will be