Because OAMB is a quadrilateral,

... the sum of its four S = 4 rt. 4 s. But OAM + ▲ OBM

...

But

=

2 rt. Ls;

BOA.

LM is supplementary to

DEF is supplementary to DEG,

and BOA = L DEG;

I. 32, Cor. 2
III. 18

I. 13 Const.

.. A LMN is equiangular to A DEF.

1. It is assumed in the proposition that the tangents at A, B, C will meet and form a triangle. Prove this.

2. Show that there may be innumerable triangles circumscribed about the © ABC equiangular to the given ▲ DEF.

3. Given a o ABC; circumscribe about it an equilateral triangle. 4. If the points of contact of the sides of the circumscribed equilateral triangle be joined, an inscribed equilateral triangle will be obtained.

5. A side of the circumscribed equilateral triangle is double of a side of the inscribed equilateral triangle, and the area of the circumscribed equilateral triangle is four times the area of the inscribed equilateral triangle.

Supply the demonstration of the proposition from the following constructions, which do not require EF to be produced : 6. In the given circle, whose centre is O, draw any diameter BOG.

Make GOA = LE, L GOC = ▲ F, and at A, B, C draw tangents intersecting at L, M, N. LMN is the required triangle. 7. At any point B on the Oce of the given circle draw a tangent PBQ, and on the tangent take any points P, Q, on opposite sides of B. At P make QPR = E, and at Q make ▲ PQR = F. Assuming that PR, QR do not touch the given circle, from 0 the centre draw perpendiculars to PR, QR, and let these perpendiculars, produced if necessary, meet the circle at A and C. At A and C draw tangents LM, LN to the circle. LMN is the required triangle.

8. In the given circle inscribe a ▲ ABC equiangular to ▲ DEF. Bisect the arcs AB, BC, CA, and at the points of bisection draw tangents.

9. Any rectilineal figure ABCDE is inscribed in a circle. Bisect the arcs AB, BC, CD, DE, EA, and at the points of bisection draw tangents. The resulting figure is equiangular to

ABCDE.

10. Two triangles are circumscribed about the o ABC, each of them equiangular to ▲ DEF; prove that they are equal in all respects.

11. Describe a triangle equiangular to a given triangle, and such that a given circle shall be touched by one of its sides, and by the other two produced. Show that there are three

it is required to inscribe a circle in ▲ ABC.

Bisects ABC, ACB by BI, CI, which intersect at I; I. 9

from I draw ID, IE, IF 1 BC, CA, AB.

I. 12

With centre I and radius ID describe a circle, which will pass through the points D, E, F.

Of this circle, ID, IE, IF will be radii;

and since BC, CA, AB are 1 ID, IE, IF,

Const.

.. BC, CA, AB will be tangents to the O DEF; III. 16 .. the DEF is inscribed in the ▲ ABC.

NOTE. This proposition is included in the more general one, to describe a circle which shall touch three given straight lines. See Appendix IV. 1, p. 250.

1. It is assumed in the proposition that the bisectors BI, CI will meet at some point I. Prove this.

2. If IA be joined, it will bisect BAC.

3. The centre of the circle inscribed in an equilateral triangle is equidistant from the three vertices.

4. The centre of the circle inscribed in an isosceles triangle is equidistant from the ends of the base.

5. Prove AF + BD + CE = FB + DC + EA = semi-perimeter

of A ABC.

6. Prove AF + BC = BD + CA = CE + AB =

of A ABC.

semi-perimeter

7. With A, B, C, the vertices of ▲ ABC as centres, describe three circles, each of which shall touch the other two.

8. Find the centre of a circle which shall cut off equal chords from the three sides of a triangle.

9. If through I a straight line be drawn || BC, and terminated by AB, AC, this parallel will be equal to the sum of the segments of AB, AC between it and BC. Examine the cases for I, I, I3, 14, in Appendix IV. 1.

10. If D, E, F, the points of contact of the inscribed circle, be joined, A DEF is acute-angled.

11. The angles of ▲ DEF are respectively complementary to half the opposite angles of ▲ ABC.

12. ABC is a triangle.

D and E are points in AB and AC, or in AB and AC produced. Prove that the vertex A, and the centres of the circles inscribed in As ABC, ADE, are collinear. 13. Draw a straight line which would bisect the angle between two straight lines which are not parallel, but which cannot be produced to meet.

Let ABC be the given triangle:

it is required to circumscribe a circle about ▲ ABC.

Bisect AB at L and AC at K;

from L and K draw LS 1 AB and KS 1 AC,

and let LS, KS intersect at S.

Join SA; and if S be not in BC, join SB, SC.

I. 10

I. 11

Const.

I. 4

.. SA, SB, SC are all equal.

With centre S and radius SA, describe a circle;

this circle will pass through the points A, B, C, and will be circumscribed about the ▲ ABC.

COR. From the three figures it appears that S, the centre of the circumscribed circle, may occupy three positions: (1) It may be inside the triangle. (2) It may be on one of the sides. (3) It may be outside the triangle.

In the first case, when S is inside the triangle, the s ABC, BCA, CAB, being in segments greater than a semicircle, are each less than a right angle;

.. the triangle is acute-angled.

III. 31

III. 31

In the second case, when S is on one of the sides as BC, L BAC, being in a semicircle, is right; .. the triangle is right-angled.

In the third case, when S is outside the triangle, ▲ BAC, being in a segment less than a semicircle, is greater than a right angle;

.. the triangle is obtuse-angled.

III. 31

And conversely, if the given triangle be acute-angled, the centre of the circumscribed circle will fall within the triangle; if the triangle be right-angled, the centre will fall on the hypotenuse; if the triangle be obtuse-angled, the centre will fall without the triangle beyond the side opposite the obtuse angle

1. It is assumed in the proposition that the perpendiculars at L and K will intersect. Prove this.

2. With which proposition in the Third Book may this proposition be regarded as identical?

3. Give an easy construction for circumscribing a circle about a right-angled triangle.

4. An isosceles triangle has its vertical angle double of each of the base angles. Prove that the diameter of its circumscribed circle is equal to the base of the triangle.

5. A quadrilateral has one pair of opposite angles supplementary. Show how to circumscribe a circle about it.

6. If a perpendicular SH be drawn from S to BC, it will bisect BC. 7. If the perpendicular in the preceding deduction meet the circle below the base at D, and above the base at E, prove

(a) L BSD = 4 CSD

= L BAC;.

(b) 4 BSE = 4 CSE = 4 ABC + ACB;

(c) LASE

=

(ABC ACB);

(d) that AD and AE bisect the interior and exterior

vertical angles at A.