Quadrilaterals. MA 341 Topics in Geometry Lecture 23

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1 Quadrilaterals MA 341 Topics in Geometry Lecture 23

2 Theorems 1. A convex quadrilateral is cyclic if and only if opposite angles are supplementary. (Circumcircle, maltitudes, anticenter) 2. A convex quadrilateral is tangential if and only if opposite sides sum to the same measure. (Incircle, incenter) 24-Oct-2011 MA

3 Bicentric Quadrilaterals A convex quadrilateral is bicentric if it is both cyclic and tangential. A bicentric quadrilateral has both a circumcircle and an incircle. A convex quadrilateral is bicentric if and only if a + c = b + d and A + C = 180 = B + D 24-Oct-2011 MA

4 Bicentric Quadrilaterals 24-Oct-2011 MA

5 Bicentric Quadrilaterals X A, X B, X C, X D points of contact of incircle and quadrilateral Bicentric iff X A X C X B X D or AX A DXC or = XB A XC C AC AX = A+ CXC BC BX +DX B D 24-Oct-2011 MA

6 Bicentric Quadrilaterals M 1, M 2, M 3, M 4 midpoints of X A X B, X B X C, X C X D, X D X A Bicentric iff M 1 M 2 M 3 M 4 is a rectangle. 24-Oct-2011 MA

7 Newton Line Theorem: (Léon Anne) Let ABCD be a quadrilateral that is not a parallelogram. P lies on the line joining the midpoints of the diagonals iff K K =K K APB CPD BPC APD 24-Oct-2011 MA

8 Proof Drop perpendicular from B & D to EF. 24-Oct-2011 MA

9 Proof E midpoint of BD, EPB= DEX (vertical angles), ΔEPB= ΔDEX (Hyp-ang) perpendiculars are equal K DPF = K BPF. 24-Oct-2011 MA

10 Likewise K APF = K CPF. Proof 24-Oct-2011 MA

11 Proof K APB = K AFB + K APF + K BPF 24-Oct-2011 MA

12 Proof K DPC = K DFC -K CPF -K DPF 24-Oct-2011 MA

13 Proof K APB = K AFB + K APF + K BPF K DPC = K DFC -K CPF K DPF So K APB + K DPC =K AFB + K DFC 24-Oct-2011 MA

14 Proof K APD = K AFD + K DPF + K APF K BPC = K BFC -K BPF K CPF So K APD + K BPC =K AFD + K CFB 24-Oct-2011 MA

15 Proof K APB + K DPC = K AFB + K DFC K APD + K BPC = K AFD + K CFB Also K AFB = K CFB (equal bases, same height) and K DFC = K AFD (equal bases, same height) So K APB + K DPC = K APD + K BPC 24-Oct-2011 MA

16 Newton Line Theorem: The center of the circle inscribed into a quadrilateral lies on the line joining the midpoints of the diagonals. 24-Oct-2011 MA

17 Proof The distance from O to each side is R. The area of each triangle then is K AOB = ½ R AB K BOC = ½ R BC K COD = ½ R CD K DOA = ½ R AD Since AB + CD = BC + AD, multiplying both sides by ½R, we get that K AOB + K COD = K BOC + K DOA Thus, O lies on the Newton Line. 24-Oct-2011 MA

18 Area A bicentric quadrilateral is a cyclic quadrilateral, so Brahmagupta s Formula applies: K = (s - a)(s -b)(s -c)(s -d) Since it is a tangential quadrilateral we know that a + c = s = b + d. Thus: s a = c, s b = d, s c = a and s d = b or K= abcd 24-Oct-2011 MA

19 k,l = tangency chords p,q = diagonals klpq K= k +l 2 2 Area 24-Oct-2011 MA

20 Area r = inradius R = circumradius 4r K 2R Oct-2011 MA

21 Inradius & Circumradius Since it is a tangential quadrilateral r K a+c Since it is bicentric, we get r abcd a+c We gain little for the circumradius 1 (ab + cd)(ac +bd)(ad +bc) R 4 abcd 24-Oct-2011 MA

22 Fuss Theorem Let x denote the distance from the incenter, I, and circumcenter, O. Then = (R - x) (R + x) r Oct-2011 MA

23 NOTE Let d denote the distance from the incenter, I, and circumcenter, O, in a triangle, then Euler s formula says or = R-d R+d r 2 2 2Rr =R -d 24-Oct-2011 MA

24 Proof Let K,L be points of tangency of incircle A + C = 180 IL = r = IK and ILC = IKA = 90 Join ΔILC and ΔIKA to form ΔCIA I A K,L C 24-Oct-2011 MA

25 Proof What do we know about ΔCIA? A + C = 180 LCI = ½C and IAK= ½A LCI + IAK = 90 CIA = 90 ΔCIA a right triangle Hypotenuse = AK+CL I A K,L C 24-Oct-2011 MA

26 Proof Area = ½r(AK+CL) = ½ AI CI or r(ak+cl) = AI CI Pythagorean Theorem gives (AK+CL) 2 = AI 2 + CI 2 From first equation we have r 2 (AI 2 + CI 2 ) = AI 2 CI = + r AI CI A I K,L C 24-Oct-2011 MA

27 Proof Extend AI and CI to meet circumcircle at F and E. DOF+ DOE = 2 DAF + 2 DCE = BAD + BCD = 180 EF is a diameter! 24-Oct-2011 MA

28 Proof X = IO = median in ΔIFE IO = (IE + IF ) - EF IO = (IE + IF ) - (2R) IE + IF 2IO 2R =2(x R ) 24-Oct-2011 MA

29 Proof From chords CE and AF, we have CI FI AI = EI AI FI = CI EI From chords CE and XY CI EI = XI YI =(R + x)(r - x) 2 2 =R -x 24-Oct-2011 MA

30 AI FI =R - x 2 2 Proof FI EI + = + AI CI CI EI CI EI EI +FI = (R - x ) (R + x ) = (R - x ) AI FI = CI EI 24-Oct-2011 MA

31 Proof (R +x ) = r (R -x ) ((R + x ) 2 +(R - x) 2 ) = (R - x ) = r (R+x) (R-x) r (R + x ) =(R - x ) x= R +r -r 4R +r 24-Oct-2011 MA

32 Other results In a bicentric quadrilateral the circumcenter, the incenter and the intersection of the diagonals are collinear. Given two concentric circles with radii R and r and distance x between their centers satisfying the condition in Fuss' theorem, there exists a convex quadrilateral inscribed in one of them and tangent to the other. 24-Oct-2011 MA

33 Other results Poncelet s Closure Theorem: If two circles, one within the other, are the incircle and the circumcircle of a bicentric quadrilateral, then every point on the circumcircle is the vertex of a bicentric quadrilateral having the same incircle and circumcircle. 24-Oct-2011 MA

DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle.

DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle. DEFINITIONS Degree A degree is the 1 th part of a straight angle. 180 Right Angle A 90 angle is called a right angle. Perpendicular Two lines are called perpendicular if they form a right angle. Congruent