PROPOSITION 3. PROBLEM. About a given circle to circumscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to circumscribe about the ○ ABC a triangle equi angular to the Construction. DEF. Produce EF both ways to G and H. At K make the BKA equal to the III. 1. DEG; I. 23. DFH. and make the BKC equal to the Through A, B, C draw LM, MN, NL perp. to KA, KB, KC. Then LMN shall be the triangle required. Proof. Because LM, MN, NL are drawn perp. to radii at their extremities. ... LM, MN, NL are tangents to the circle. III. 16. And because the four angles of the quadrilateral AKBM together four rt. angles; = = two rt. angles. I. 32. Cor. and of these, the 2 KAM, KBM are rt. angles; Constr. .. the 2 AKB, AMB together But the DEG, DEF together the AKB, AMB = the and of these, the AKB = the the AMB the = = two rt. angles; I. 13. DEF. Constr. Similarly it may be shewn that the LNM the DFE. the ALMN is equiangular to the ▲ DEF, and it is circumscribed about the ABC. Q.E.F. PROPOSITION 4. PROBLEM. To inscribe a circle in a given triangle. E B F Let ABC be the given triangle. It is required to inscribe a circle in the ▲ ABC. Construction. Bisect the ABC, ACB by the st. lines I. 9. I. 12. BI, Cl, which intersect at I. From draw IE, IF, IG perp. to AB, BC, CA. Similarly it may be shewn that IF = IG. .. IE, IF, IG are all equal. With centre 1, and radius IE, describe a circle. For since IE, IF, IG, being equal, are radii of the O EFG ; and since the * at E, F, G are rt. angles; Constr. EFG is touched at these points by AB, BC, CA: .. the .. the EFG is inscribed in the A ABC. III. 16. Q.E.F. NOTE. From page 111 it is seen that if Al is joined, then Al bisects the angle BAČ: hence it follows that The bisectors of the angles of a triangle are concurrent, the point of intersection being the centre of the inscribed circle. The centre of the circle inscribed in a triangle is usually called its in-centre. DEFINITION. A circle which touches one side of a triangle and the other two sides produced is said to be an escribed circle of the triangle. To draw an escribed circle of a given triangle. Let ABC be the given triangle, of which the two sides AB, AC are produced to E It is required to describe a circle touching Bisect the CBE, BCF by the st. I. 12. Then in the As I,BG, I,BH, H and the GB=the IHB, Because being rt. angles; E also B is common; Similarly it may be shewn that I1H=I1K; .. IG, H, I,K are all equal. With centre 1, and radius IG, describe a circle. This circle must pass through the points G, H, K; For since IH, IG, IK, being equal, are radii of the O HGK, :. the GHK is touched at these points by BC, and by AB, AC produced: ..the GHK is an escribed circle of the ▲ ABC. Q.E. F. It is clear that every triangle has three escribed circles. NOTE. From page 112 it is seen that if Al, is joined, then Al, bisects the angle BAČ: hence it follows that The bisectors of two exterior angles of a triangle and the bisector of the third angle are concurrent, the point of intersection being the centre of an escribed circle. B PROPOSITION 5. PROBLEM. To circumscribe a circle about a given triangle. E B C Let ABC be the given triangle. It is required to circumscribe a circle about the ABC. Construction. Draw DS bisecting AB at rt. angles; I. 11. and draw ES bisecting AC at rt. angles. Then since AB, AC are neither par', nor in the same st. line, DS and ES must meet at some point S. Proof. Join SA; and if S be not in BC, join SB, SC. Then in the ▲ ADS, BDS, AD = BD, Because and DS is common to both; and the ADS the ... SASB. BDS, being rt. angles; I. 4. Similarly it may be shewn that SC = SA. ... SA, SB, SC are all equal. With centre S, and radius SA, describe a circle: this circle must pass through the points A, B, C, and is therefore circumscribed about the A ABC. It follows that Q.E.F. (i) when the centre of the circumscribed circle falls within the triangle, each of its angles must be acute, for cach angle is then in a segment greater than a semicircle : (ii) when the centre falls on one of the sides of the triangle, the angle opposite to this side must be a right angle, for it is the angle in a semicircle: (iii) when the centre falls without the triangle, the angle opposite to the side beyond which the centre falls, must be obtuse, for it is the angle in a segment less than a semicircle. Therefore, conversely, if the given triangle be acute-angled, the centre of the circumscribed circle falls within it: if it be a rightangled triangle, the centre falls on the hypotenuse: if it be an obtuse-angled triangle, the centre falls without the triangle. NOTE. From page 111 it is seen that if S is joined to the middle point of BC, then the joining line is perpendicular to BC. Hence the perpendiculars drawn to the sides of a triangle from their middle points are concurrent, the point of intersection being the centre of the circle circumscribed about the triangle. The centre of the circle circumscribed about a triangle is usually called its circum-centre. EXERCISES. ON THE INSCRIBED, CIRCUMSCRIBED, AND ESCRIBED CIRCLES OF A 1. TRIANGLE. An equilateral triangle is inscribed in a circle, and tangents are drawn at its vertices, prove that (i) the resulting figure is an equilateral triangle : (ii) its area is four times that of the given triangle. 2. Describe a circle to touch two parallel straight lines and a third straight line which meets them. can be drawn, and that they are equal. Shew that two such circles 3. Triangles which have equal bases and equal vertical angles have equal circumscribed circles. 4. I is the centre of the circle inscribed in the triangle ABC, and is the centre of the circle which touches BC and AB, AC produced : shew that A, I, I are collinear. 5. If the inscribed and circumscribed circles of a triangle are concentric, shew that the triangle is equilateral; and that the diameter of the circumscribed circle is double that of the inscribed circle. 6. ABC is a triangle, and I, S are the centres of the inscribed and circumscribed circles; if A, I, S are collinear, shew that AB=AC |