Join

two opposite angles as A, D;
and about ADI describe a cir-
cle meeting GA, GD produced
if necessary, in K, L.
FC, KI, LI, LK. Then be-
cause AKID is a quadrilateral
figure inscribed in a circle, the
angle AKI is equal to ADE:
and for the same reason ADE
is equal to GFE, .. the angle
AKI is equal to GFE, and KI
is parallel to FH. In the same

H

manner it may be shewn that the angle AIL is equal to ADL and consequently to CBI, and LI parallel to HB. Again the angle KLD is equal to DAF, i. e. to FCD, and KL is parallel to FC.

Hence GF: FC :: GK: KL,

and FC FH ::

KL : KI,

whence G, H, I will be in a straight line.

SECT. VII.

(1.) THE vertical angle of any oblique-angled triangle inscribed in a circle, is greater or less than a right angle, by the angle contained by the base and the diameter drawn from the extremity of the base.

Let ABC be a triangle inscribed in a circle. From A draw the diameter AD; join BD; the angle ABC

is

greater or less than a right angle, by the angle CAD.

For the angle ABD in a semicircle is a right angle; and ABC is equal to the

sum of ABD and DBC in one case; and is equal to their difference in the other; and in each case_DBC= DAC in the same segment.

(2.) If from the vertex of an isosceles triangle a circle be described with a radius less than one of the equal sides, but greater than the perpendicular; the parts of the base cut off by it, will be equal.

From the vertex O of the isosceles triangle AOB, with a radius less than 40, but greater than the perpendicular from O on AB, let a

D

E

circle be described, cutting AB in C'and D; AC=BD. Join EF, OC, OD.

and. EF is parallel to AB; and (ii. 1.) the arc FC equal to the arc DE, or the angle FOC=DOE; but AO, OC are equal to BO, OD each to each; .. AC= DB.

(3.) If a circle be inscribed in a right-angled triangle; the difference between the two sides containing the right angle and the hypothenuse, is equal to the diameter of the circle.

GG

Let DEF be a circle, inscribed in the right-angled triangle ABC. The difference between AC, and AB, BC, is equal to the diameter of the circle.

Find O the centre, and join OD, QE. Then the angles at D, B, and E being

right angles, and OD=OE, OB is a square; and DB, BE are equal to OD, OE, i. e. are together equal to the diameter of the circle. Now (Eucl. iii. 36. Cor.) CE= CF, and AD=AF; i. e. AC is equal to AD and CE; whence it is less than the sides containing the right angle, by DB and BE, or by the diameter of the circle..

(4.) If a semicircle be inscribed in a right-angled triangle, so as to touch the hypothenuse and perpendicular, and from the extremity of its diameter a line be drawn through the point of contact to meet the perpendicular produced; the part produced will be equal to the perpendicular.

Let the semicircle ADE touch the hypothenuse BC of the right-angled triangle ABC in D, and the perpendicular in A; and from E let ED be drawn to meet AB produced in F; AB=BF.

Join AD. Since ADE is a right an

B

gle, ADF is also a right angle, and .. equal to DAF, DFA together. But DAF is equal to BDA, since BD

= BA, being tangents from the same point B without the circle; and .. the angle BFD=BDF, and BF= BD=BA.

(5.) If the base of any triangle be bisected by the diameter of its circumscribing circle, and from the extremity of that diameter a perpendicular be let fall upon the longer side; it will divide that side into segments, one of which will be equal to half the sum, and the other to half the difference of the sides.

Let the base BC of the triangle ABC be bisected in E, by the diameter of the circumscribing circle ACD; and from D draw DF perpendicular to AB the longer side; BF will be equal to half the sum, and AF to half the difference of AB, BC.

E

H

Join DA, DB, DC; and make BG = BC; join DG. Since BG= BC, and BD is common, and the angle GBD=CBD, since the arc AD=DC; .. DG= DC= DA, and DF is at right angles to AG, .. AF= FG. Whence the sum of AB and BC is equal to AG and 2 BG, i. e. to 2 BF; and the difference of AB and BC is equal to the difference of AB and BG, i. e. to 2 AF.

(6.) The same supposition being made, as in the last proposition; if from the point, where the perpendicular meets the longer side, another perpendicular be let fall on the line bisecting the vertical angle; it will pass through the middle of the base..

The same construction being made as before; (sce last Fig.) let FH be drawn perpendicular to BD, which bisects the vertical angle; FH will pass through E.

Because CB=BG, and the angle CBD=GBD,

.. BD is perpendicular to CG; and .. FH is parallel to CG. But since AF FG, .. (Eucl. vi. 2.) AC is bisected by FH; which .. passes through E.

(7.) If a point be taken without a circle, and from it tangents be drawn to the circle, and another point be taken in the circumference between the two tangents, and a tangent be drawn to it; the sum of the sides of the triangle thus formed is equal to the sum of the two tangents.

From a given point D let two tangents DA, DB be drawn; and to C any point in the circumference between them, let a tangent ECF be drawn. The sum of the sides of the triangle is equal to the two tangents DA and DB.

=

E

B

Since AE EC, and FC = FB, .. DE, EF, FD together are equal to AD and DB together. In the same manner, if through any other point in the arc ACB a tangent be drawn, it will be equal to the two segments of DA, DB intercepted between it, and the points of contact A and B ; and the three sides of the triangle so formed will be equal to DA, and DB together.

(8.) Of all triangles on the same base and between the same parallels, the isosceles has the greatest vertical angle.

Let ABC be an isosceles triangle on the base AC, and between the parallels AC, BD. It has a greater vertical angle

B