PROPOSITION 2. PROBLEM. In a given circle to inscribe a triangle equiangular to a given triangle. G Let ABC be the given circle, and DEF the given triangle. It is required to inscribe in the O ABC a triangle equiangular to the A DEF. Construction. At any point A, on the Oce of the O ABC, draw the tangent GAH. III. 17. At A make the L GAB equal to the 2 DFE; I. 23. and make the 2 HAC equal to the 2 DEF. I. 23. Join BC. Then ABC shall be the triangle required. Proof. Because GH is a tangent to the O ABC, and from A its point of contact the chord AB is drawn, .. the 2 GAB= the L ACB in the alt. segment : III. 32. but the į GAB= the L DFE; Constr. ... the ACB=the į DFE. Similarly the 2 HAC = the 2 ABC, in the alt. segment : ... the L ABC = the % DEF. Constr. Hence the third – BAC = the third – EDF, for the three angles in each triangle are together equal to two rt. angles. I. 32. ::: the ABC is equiangular to the A DEF, and it is inscribed in the O ABC. Q E.F. = PROPOSITION 3. PROBLEM. About a given circle to circumscribe a triangle equiangular to a given triangle. III. 1. Let ABC be the given circle, and DEF the given triangle. Find K the centre of the O ABC, radius KB. I. 23. 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 ; I. 32. Cor. and of these, the 2' KAM, KBM are rt. angles; Constr. .. the 2" AKB, AMB together=two rt. angles. But the 'DEG, DEF together = two rt. angles; I. 13. ... the _AKB, AMB = the < DEG, DEF ; and of these, the L AKB= the _ DEG; Constr. ... the L AMB= the < DEF. Similarly it may be shewn that the < LNM= the 2 DFE. .. the third MLN=the third – EDF. the A LMN is equiangular to the A DEF, and it is circumscribed about the O ABC. Q.E.F. I. 32. PROPOSITION 4. PROBLEM. E B F Construction. Bisect the < ABC, ACB by the st. lines BI, CI, which intersect at I. I. 9. From | draw IE, IF, IG perp. to AB, BC, CA. 1. 12. Proof. Then in the As EIB, FIB, the < EBI = the L FBI; Constr. Because and the 2 BEI = the 2 BFI, being rt. angles ; and Bl is common ; I. 26, Similarly it may be shewn that IF = IG. and it will be inscribed in the A ABC. For since IE, IF, IG, being equal, are radii of the O EFG; and since the 4 at E, F, G are rt. angles; Constr. .. the O EFG is touched at these points by AB, BC, CA: III. 16. ... the © EFG is inscribed in the A ABC. Q.E.F. a Note. From page 111 it is seen that if Al is joined, then Al bisects the angle BAC : 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 and F. It is required to describe a circle touching BC, and AB, AC produced. Bisect the 48 CBE, BCF by the st. lines Blı, Cly, which intersect at 11. I. 9. H C 1. 12. K (the 21_BG=the LI,BH, Constr. and the L 1 GB=the LHB, Because being rt. angles ; F also 1, B is common ; E :: IG=1H. Similarly it may be shewn that I, H=1,K; :: ,,G, H, I,K are all equal. With centre l, and radius I,G, describe a circle. This circle must pass through the points G, H, K; and it will be an escribed circle of the A ABC. For since 1, H, I,G, I,K, being equal, are radii of the O HGK, and since the angles at H, G, K are rt. angles, :. the O GHK is touched at these points by BC, and by AB, AC produced : :: the O 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 Ali 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. PROPOSITION 5. PROBLEM. a Let ABC be the given triangle. 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. Join SA; Then in the A8 ADS, BDS, AD= BD, Because and DS is common to both; and the 2 ADS = the 2 BDS, being rt. angles ; .. SA=SB. 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. Q.E.F. It follows that (i) when the centre of the circumscribed circle falls within the triangle, each of its angles must be acute, for each angle is then in a segment greater than a semicircle : (ii) when the centre falls on one of the ides the triangle, the angle opposite to this side must be a right angle, for it is the angle in a semicircle : а |