MA 341 – Topics in Geometry Lecture 21
Transcript of MA 341 – Topics in Geometry Lecture 21
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QuadrilateralsQuadrilaterals
MA 341 – Topics in GeometryLecture 21
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Ptolemy’s TheoremPtolemy s TheoremLet a, b, c, and d be the l h f
b
lengths of consecutive sides of a cyclic quadrilateral and let x a
y
quadrilateral and let x and y be the lengths of the diagonals. Then
cx
the diagonals. Thenac + bd = xy. d
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Ptolemy’s TheoremPtolemy s TheoremWe have: ∆ABP ~ ∆CDP & ∆ ∆
bBC
∆BCP ~ ∆DAP.So
a
u s
r Pb c
rv
A
a u r b u sandc s v d r v
as = uc, br = ud, and uv = rs.sa2 + rb2 = uac+ubd = u(ac+bd)
d
D
xu2+xrs = xu2+xuv=xu(u+v)=uxyBy Stewart’s Theorem
2 2 2( bd) s b s17-Oct-2011 MA 341 001 3
2 2 2u(ac bd) sa rb xu xrs uxyac bd xy
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The Converse of Ptolemy’s Theorem
Let a, b, c, and d be the lengths of consecutive sides of a quadrilateral and let x and y be the lengths of the diagonals. If
ac + bd = xy,then the quadrilateral is a cyclic quadrilateralthen the quadrilateral is a cyclic quadrilateral.
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Euler’s TheoremEuler s TheoremLet a, b, c, and d be the l h f
b
lengths of consecutive sides of a quadrilateral, m and n lengths of a
p
m and n lengths of diagonals, and x the distance between
ac
x
qdistance between midpoints of diagonals. Then d
q
a2 + b2 + c2 + d2 = m2 + n2 + 4x2
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Brahmagupta’s TheoremBrahmagupta s TheoremThere is an analog of H ’s F l f
b
Heron’s Formula for special quadrilaterals.
a
Let a, b, c, and d be lengths of consecutive
c
gsides of cyclic quadrilateral, then
d
( )( )( )( )K s a s b s c s d
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Area of a QuadrilateralArea of a Quadrilateralc
Using triangle trigonometryDC
d
Using triangle trigonometryyou can show that q
b
A
θp
a
B
2 2 2 21 1sin ( ) tan2 41
A pq b d a c
2 2 2 2 2 2 2
1
1 4 ( )4
p q b d a c
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12( )( )( )( ) cos ( )s a s b s c s d abcd A C
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MaltitudesMaltitudesFor a quadrilateral the maltitude( id i t ltit d ) i di l (midpoint altitude) is a perpendicular through the midpoint of one side
di l t th it idperpendicular to the opposite side.
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MaltitudesMaltitudesRectangle
Square – same
Parallelogram
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MaltitudesMaltitudesRhombus
Kite
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MaltitudesMaltitudesTrapezoid
Isosceles trapezoidIsosceles trapezoid
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MaltitudesMaltitudesCyclic quadrilaterals
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Quadrilaterals and CirclesQuadrilaterals and Circles
• For a cyclic quadrilateral area is easy For a cyclic quadrilateral, area is easy and there are nice relationships
• Maybe maltitudes of cyclic quadrilateral • Maybe maltitudes of cyclic quadrilateral are concurrentC t ll h d il t l i • Can we tell when a quadrilateral is cyclic?
• Can we tell when a quadrilateral has an inscribed circle?
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TheoremTheorem
For a cyclic quadrilateral the maltitudesFor a cyclic quadrilateral the maltitudesintersect in a single point, called the anti-centercenter.
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ProofProof
Let O = center of circleLet O = center of circleG = centroid,
intersection of midlinesintersection of midlines
Let T = point on ray OG so that OG = GT
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ProofProof
PG = RGPG = RGOG = GTPGT RG0PGT = RG0∆PGT = ∆RG0
PTG = ROGPTG = ROG
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ProofProof
OR CDOR CDThus, PT CDT li ltit dT lies on maltitudethrough P
Use ∆RGT = ∆PG0to show T lies on maltitude through R g
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ProofProof
Using other midline we show T lies on Using other midline we show T lies on maltitudes through Q and S.
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AnticenterAnticenterT = anticenter = i t ti f intersection of maltitudesG = midpointO = circumcenter
OG = GTOG = GT
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Other Anticenter PropertiesOther Anticenter PropertiesPerpendiculars from
id i t f midpoint of one diagonal to other i t t t Tintersect at T.
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Other Anticenter PropertiesOther Anticenter PropertiesConstruct 9-point i l f th f circles of the four
triangles ∆ABD,∆BCD, ∆ABC, and ,∆ADC.The 4 circles The 4 circles intersect at the anticenteranticenter.17-Oct-2011 MA 341 001 21
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Other Anticenter PropertiesOther Anticenter PropertiesThe centers of the 9 i t i l 9-point circles are concyclic with
t Tcenter T.
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