270 PROPOSITION 2. PROBLEM. In a given circle to inscribe a triangle equiangular to a given triangle. w H Let ABC be the given circle, and DEF the given triangle. It is required to inscribe in the ○ ABC a triangle equiangular to the A DEF. Join BC. Proof. Then ABC shall be the triangle required. Because GH is a tangent to the ABC, and from A its point of contact the chord AB is drawn, .. the GAB=the ▲ ACB in the alt. segment: III. 32. but the GAB = the .. the ACB = the DFE; Constr. DFE. Similarly the HAC = the ..the ABC the DEF. = 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 ▲ DEF, and it is inscribed in the ABC. QE.F. 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 equiangular to the DEF. Construction. Produce EF both ways to G and H. and draw any radius KB. III. 1. DEG; I. 23. At K make the BKA equal to the 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. Because LM, MN, NL are drawn perp. to radii at their extremities. Proof. ... 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 AKB, AMB together two rt. angles. two rt. angles; I. 13. 3 DEG, DEF; Similarly it may be shewn that the LNM = the DFE. I. 32. .. the third MLN the third EDF. .. the LMN 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 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 BI, Cl, which intersect at I. From draw IE, IF, IG perp. to AB, BC, CA. I. 9. I. 12. Similarly it may be shewn that IFIG. .. IE, IF, IG are all equal. With centre I, and radius IE, describe a circle. For since IE, IF, IG, being equal, are radii of the EFG; and since the at E, F, G are rt. angles; Constr. .. the EFG is touched at these points by AB, BC, CA: ... the EFG is inscribed in the ▲ 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 8 CBE, BCF by the st. Because I. 12. Then in the 3 I,BG, I,BH, also B is common; .. IG=I2H. Similarly it may be shewn that I1H=I1K; Ε B .. IG, H, I1K are all equal. With centre l, and radius IG, describe a circle. K HGK, This circle must pass through the points G, H, K; and it will be an escribed circle of the ▲ ABC. For since IH, IG, IK, being equal, are radii of the and since the angles at Ĥ, G, K are rt. angles, the GHK is touched at these points by BC, and by AB, AC produced : Q. E. F. the GHK is an escribed circle of the AABC. It is clear that every triangle has three escribed circles. NOTE. From page 112 it is seen that if Al, is joined, then Al1 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. To circumscribe a circle about a given triangle. 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, 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 each 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: |