than any other triangle ADC on the same base, and between the same parallels.

About ABC describe a segment of a circle ABC; and since B is the middle point of the arc, and BD is parallel to AC, BD is a tangent at B. Let the arc cut AD in E; join EC. Then the angle ABC= AEC, and.. is greater than ADC.

COR. Of all triangles on the same base and having the same vertical angle, the isosceles is the greatest. For the triangle AEC has the same vertical angle with ABC, and ABC= ADC on the same base and between the same parallels; but ADC is greater than AEC, :. ABC is greater than AEC.

(9.) If through the vertex of an equilateral triangle a perpendicular be drawn to the side, meeting a perpendicular to the base drawn from its extremity; the line intercepted between the vertex and the latter perpendicular is equal to the radius of the circumscribing circle.

Let BE perpendicular to AB meet AE, which is perpendicular to the base AC, in E; BE is equal to the radius of the circle described about ABC.

E

H G

Draw BF, CG perpendiculars to the sides; and produce CG to H. Then CI is equal to the radius of the circle described about ABC; and EBIH is a parallelogram. And since CF is equal to FA, (Eucl. vi. 2.) CI is equal to IH, i. e. to the opposite side BE; and .. BE is equal to the radius of the circumscribing circle.

...

(10.) If a triangle be inscribed in a semicircle, and a perpendicular drawn from any point in the diameter, meeting one side, the circumference, and the other side produced; the segments cut off will be in continued proportion.

Let ABC be a triangle in the semicircle ABC; and from any point D in the diameter, let DF be drawn perpendicular to AD, meeting BC, the circumference, and AB produced, in E, G, F; DE: DG :: DG : DF.

F

D

G

E

For the angles at E being equal, and the angles at Bright angles, .. the angle ECD is equal to BFD; and the angles at D are right angles; .. the triangles EDC, ADF are similar, and therefore

DF: DA:: DC: DE,

but DA: DG :: DG: DC; .. ex æquo DF: DG :: DG: DE.

(11.) If a triangle be inscribed in a semicircle, and one side be equal to the semi-diameter; the other side will be a mean proportional between that side and a line equal to that side and the diameter together.

Let ABC be a triangle inscribed in the semicircle, and let BC be equal to the semi-diameter; then A will

BC BA: BA : BC+CA.

Produce AC to D, making CD equal to the semidiameter. Take O the centre. Join BD, BO. Since

BO= BC, the angle BCO is equal to BOC, i. e. to OAB and OBA together, or to 2 BAC. But BCA is equal to CBD and CDB together, i. e. to 2 CDB, since CB= CD; hence the angle BAC-BDC, and _BA=BD; also the triangles BAD, BCD are similar;

.. BC : (BD=) BA :: BA : AD, which is equal to BC and CA together.

(12.) If a circle be inscribed in a right-angled triangle; to determine the least angle that can be formed by two lines drawn from the extremity of the hypothenuse to the circumference of the circle.

Let ABC be a right-angled

triangle, in which a circle DEG is inscribed. On AC describe a segment of a circle ADC, which may touch the inscribed circle in

G

some point, as D. The lines AD, DC, drawn to this point from A and C, contain an angle less than the lines drawn to any other point in the circumference of the circle DEG.

For take any other point E, and join AE, EC; produce CE to F, and join AF. The exterior angle AEC is greater than AFC, i. e. than ADC, which is in the same segment. And the same may be proved of lines drawn to every other point in the circumference of the circle DEG.

(13.) If an equilateral triangle be inscribed in a

circle, and through the angular points another be circumscribed; to determine the ratio which they bear to each other.

Let ABC be an equilateral triangle inscribed in the circle, about which another DEF is circumscribed, touching the circle in the points A, B, C.

Since DA touches the circle, the P

=

B

angle DAB ACB (Eucl. iii. 32.); but ACB=ABC; .. DAB=ABC, and they are alternate angles, .. DF is parallel to BC. In the same manner it may be shewn that AB is parallel to FE, .. ABCF is a parallelogram ; and the triangle ABC is equal to AFC. In the same manner ABC may be shewn to be equal to each of the triangles ABD, BCE; and .. it is one fourth of the circumscribing triangle.

(14.) A straight line drawn from the vertex of an equilateral triangle inscribed in a circle to any point in the opposite circumference, is equal to the two lines together, which are drawn from the extremities of the base to the same point.

Let ABC be an equilateral triangle inscribed in a circle; from B draw BD to any point D in the circumference. Join AD, CD. BD is equal to AD and CD together.

Make DE=DA, and join AE. The

B

angle DAE is equal to the angle DEA; but ADE = ACB in the same segment, .. DAE and DEA together are equal to CBA and CAB together; whence DAE

=CAB; and taking away the common angle CAE, DAC = EAB; but DCA = EBA, and AC=AB, .. BE=DC; and BD is equal to AD and CD together.

(15.) If the base of a triangle be produced both ways so that each part produced may be equal to the adjacent side, and through the extremities of the parts produced and the vertex a circle be described; the line joining its centre and the vertex of the triangle will bisect the angle at the vertex.

Let AC a side of the triangle ABC be produced both ways till AD =AB, and CE = CB; and through D, B, E let a circle be described,

whose centre is O. If OB be joined, it will bisect the angle ABC.

Join BD, BE, OA, OD, OE. Since DA=AB, the angle ABD is equal to ADB; but the angle OBD is equal to ODB, and .. the angle OBA is equal to ODA. In the same manner it may be shewn that the angle OBC is equal to OEC; and since ODA is equal to OEC, OBA is equal to OBC; or ABC is bisected by OB.

(16.) If an isosceles triangle be inscribed in a circle, and from the vertical angle a line be drawn meeting the circumference and the base; either equal side is a mean proportional between the segments of the line thus drawn.

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