Sinces DLK, KLI, &c. = each other, ... arcs DK, KI, &c. =

And..

No. of arcs in DC =

No. of arcs in CB =

No. of arcs in DC

each other;

No. ofs in DLC
No. ofs in CLB'

Prop. 58.

No. of arcs in CB

Arc DC

DLC

i. e.

or Arc DC: Arc CB::

Wherefore in the same, &c.

DLC : / CLB.

Since angles at the centre of a circle vary as the contained arcs upon which they stand; it is frequently necessary for practical purposes to consider an arc as the measure of an angle; or the angle to be measured by the number of equal units or degrees contained in its corresponding arc, the whole circumference of the circle being usually divided into 360 degrees.

The angle made by the intersection of two straight lines within a circle is measured by half the sum of the two contained arcs.

Let str. lines BE and CD cut each other in pt. A, within the BCFD; then will half the sum of the arcs DE and BC be the measure of the of intersection BAC.

COR.-Errors of "Excentricity" in instruments for the measurement of angles, are corrected by taking the half sum, or mean, between the readings of two opposite Verniers.

PROP. LXXVIII. THEOR.

If an angle of a triangle be bisected by a straight line which cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the bisecting line.

Let ABC be a , and the str. line AD bisect the BAC; then shall BA. AC= BD. DC + AD2.

PROP. LXXIX. THEOR.

If from any angle of a triangle a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

Let ABC be a A, and AD 1 BC; AE the diam. of the circumscribing

then shall BA. AC=
= AD. AE.

ABEC;

rt. / BDA = ECA in semi-O, and ABD=/ AEC,in same seg.; As ABD, AEC are equiangular; BA: AD: EA: AC;

Prop. 59.

Prop. 56.

Then ...

Prop. 73.

Prop. 65.

PROP. LXXX. THEOR.

The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles together, contained by its opposite sides.

Let ABCD be any quadrilateral fig. inscrib. in a ; draw the diagonals AC, BD: then shall AC. BD = AB.CD + AD. BC.

=

B

E

< BCE,

Prop. 56.

.. As ABD, BCE, are equiangular,
.. BC: CE :: BD: AD;

CE. BDAD. BC.

Prop. 73.

Prop. 65.

but AE. BD+CE.BDAC.BD

.. AE.BD+CE. BD-AB. CD+AD. BC

AC.BD=AB.CD+AD. BC.

Prop. 41.

Therefore the rectangle, &c.

PROP. LXXXI. THEOR.

Equiangular triangles have their bases in the same ratio as their altitudes, or perpendiculars upon the bases from the opposite and equal angles.

Let the As ABC, DEF have the s A, B, C of the one ▲, respectively equal to the Ls D, E, F of the other; draw AG BC and DHEF; then shall DH: AG :: EF : BC; or DH EF:: AG: BC.