MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001...
Transcript of MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001...
![Page 1: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/1.jpg)
Quadrilaterals
MA 341 – Topics in GeometryLecture 23
![Page 2: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/2.jpg)
Theorems1. 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 341 001 2
![Page 3: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/3.jpg)
Bicentric QuadrilateralsA 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 341 001 3
![Page 4: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/4.jpg)
Bicentric Quadrilaterals
24-Oct-2011 MA 341 001 4
![Page 5: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/5.jpg)
Bicentric Quadrilaterals
24-Oct-2011 MA 341 001 5
XA, XB, XC, XD points of contact of incircle and quadrilateralBicentric iffXAXC XBXD or
or CA
A C
DXAX =X B X C
A C
B D
AX +CXAC =BC BX +DX
![Page 6: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/6.jpg)
Bicentric Quadrilaterals
24-Oct-2011 MA 341 001 6
M1, M2, M3, M4midpoints of XAXB, XBXC, XCXD, XDXA
Bicentric iffM1M2M3M4 is a rectangle.
![Page 7: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/7.jpg)
Newton Line
24-Oct-2011 MA 341 001 7
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
APB CPD BPC APDK K =K K
![Page 8: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/8.jpg)
Proof
24-Oct-2011 MA 341 001 8
Drop perpendicular from B & D to EF.
![Page 9: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/9.jpg)
Proof
24-Oct-2011 MA 341 001 9
E midpoint of BD, EPB= DEX (vertical angles),ΔEPB= ΔDEX (Hyp-ang) perpendiculars are equalKDPF = KBPF.
![Page 10: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/10.jpg)
Proof
24-Oct-2011 MA 341 001 10
Likewise KAPF = KCPF.
![Page 11: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/11.jpg)
Proof
24-Oct-2011 MA 341 001 11
KAPB = KAFB + KAPF + KBPF
![Page 12: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/12.jpg)
Proof
24-Oct-2011 MA 341 001 12
KDPC = KDFC - KCPF - KDPF
![Page 13: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/13.jpg)
Proof
24-Oct-2011 MA 341 001 13
KAPB = KAFB + KAPF + KBPF
KDPC = KDFC - KCPF – KDPF
SoKAPB + KDPC = KAFB + KDFC
![Page 14: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/14.jpg)
Proof
24-Oct-2011 MA 341 001 14
KAPD = KAFD + KDPF + KAPF
KBPC = KBFC - KBPF – KCPF
SoKAPD + KBPC = KAFD + KCFB
![Page 15: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/15.jpg)
Proof
24-Oct-2011 MA 341 001 15
KAPB + KDPC = KAFB + KDFCKAPD + KBPC = KAFD + KCFBAlso KAFB = KCFB (equal bases, same height) and KDFC = KAFD (equal bases, same height)
So KAPB + KDPC = KAPD + KBPC
![Page 16: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/16.jpg)
Newton Line
24-Oct-2011 MA 341 001 16
Theorem: The center of the circle inscribed into a quadrilateral lies on the line joining the midpoints of the diagonals.
![Page 17: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/17.jpg)
Proof
24-Oct-2011 MA 341 001 17
The distance from O to each side is R.The area of each triangle then isKAOB = ½ R|AB| KBOC = ½ R|BC|KCOD = ½ R|CD| KDOA = ½ R|AD|Since AB + CD = BC + AD, multiplying both
sides by ½R, we get that KAOB + KCOD= KBOC + KDOA
Thus, O lies on the Newton Line.
![Page 18: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/18.jpg)
Area
24-Oct-2011 MA 341 001 18
A bicentric quadrilateral is a cyclic quadrilateral, so Brahmagupta’s Formula applies:
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 = (s - a)(s -b)(s -c)(s -d)
K = abcd
![Page 19: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/19.jpg)
Area
24-Oct-2011 MA 341 001 19
k,l = tangency chordsp,q = diagonals
2 2klpqK =k +l
![Page 20: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/20.jpg)
Area
24-Oct-2011 MA 341 001 20
r = inradiusR = circumradius
2 24r K 2R
![Page 21: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/21.jpg)
Inradius & Circumradius
24-Oct-2011 MA 341 001 21
Since it is a tangential quadrilateral
Since it is bicentric, we get
We gain little for the circumradius
Kr
a +c
abcdra +c
1 (ab+cd)(ac+bd)(ad+bc)R4 abcd
![Page 22: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/22.jpg)
Fuss’ Theorem
24-Oct-2011 MA 341 001 22
Let x denote the distance from the incenter, I, and circumcenter, O. Then
2 2 21 1 1+ =
(R -x) (R +x) r
![Page 23: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/23.jpg)
NOTE
24-Oct-2011 MA 341 001 23
Let d denote the distance from the incenter, I, and circumcenter, O, in a triangle, then Euler’s formula says
or
1 1 1+ =R -d R +d r
2 22Rr =R -d
![Page 24: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/24.jpg)
Proof
24-Oct-2011 MA 341 001 24
Let K,L be points of tangency of incircleA + C = 180IL = r = IK andILC = IKA = 90Join ΔILC and ΔIKA to formΔCIA
K,LA C
I
![Page 25: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/25.jpg)
Proof
24-Oct-2011 MA 341 001 25
What do we know about ΔCIA?A + C = 180LCI = ½C and IAK= ½ALCI + IAK = 90CIA = 90ΔCIA a right triangleHypotenuse = AK+CL K,LA C
I
![Page 26: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/26.jpg)
Proof
24-Oct-2011 MA 341 001 26
Area = ½r(AK+CL) = ½ AI CI orr(AK+CL) = AI CI
Pythagorean Theorem gives(AK+CL)2 = AI2 + CI2
From first equation we haver2(AI2 + CI2) = AI2 CI2
K,LA C
I
2 2 21 1 1= +r AI CI
![Page 27: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/27.jpg)
Proof
24-Oct-2011 MA 341 001 27
Extend AI and CI to meet circumcircle at F and E.DOF+DOE =
2DAF + 2DCE= BAD + BCD= 180
EF is a diameter!
![Page 28: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/28.jpg)
Proof
24-Oct-2011 MA 341 001 28
X = IO = median in ΔIFE.2 2 2 21 1IO = (IE +IF ) - EF
2 4
2 2=2(x R )
2 2 2 21 1IO = (IE +IF ) - (2R)2 4
2 2 2 2IE +IF 2IO 2R
![Page 29: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/29.jpg)
Proof
24-Oct-2011 MA 341 001 29
From chords CE and AF, we have
From chords CE and XY
CI AI=FI EI
CI EI =XI YI
AI FI =CI EI
=(R +x)(R -x)2 2=R -x
![Page 30: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/30.jpg)
Proof
24-Oct-2011 MA 341 001 30
2 2
2 2 2 2 2 21 1 FI EI+ = +
AI CI CI EI CI EI
2 2AI FI =R -x
2 2
2 2 2EI +FI=(R -x )
2 2
2 2 22(R +x )=(R -x )
AI FI =CI EI
![Page 31: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/31.jpg)
Proof
24-Oct-2011 MA 341 001 31
)2 2
2 2 2((R +x +(R -x) )=
(R -x )
2 2 2 2 2 22r (R +x ) =(R -x )2 2 2 2x = R +r -r 4R +r
2 2
2 2 2 21 2(R +x )=r (R -x )
2 2 21 1 1= +r (R +x) (R -x)
![Page 32: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/32.jpg)
Other results
24-Oct-2011 MA 341 001 32
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.
![Page 33: MA 341 – Topics in Geometry Lecture 23droyster/courses/fall11/MA341... · 24-Oct-2011 MA 341 001 14 K APD = K AFD + K DPF + K APF K BPC = K BFC-K BPF –K CPF So K APD + K BPC =K](https://reader033.fdocuments.us/reader033/viewer/2022042310/5ed805b1cba89e334c671848/html5/thumbnails/33.jpg)
Other results
24-Oct-2011 MA 341 001 33
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.