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9/20/2011 1 Ceva’s Theorem MA 341 – Topics in Geometry Lecture 11 Ceva’s Theorem The three lines containing the vertices A, B, and C of ᇞABC and intersecting opposite sides at points L, M, and N, respectively, are concurrent if and only if 2 21-Sept-2011 MA 341 001 Ceva’s Theorem 3 21-Sept-2011 MA 341 001

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Page 1: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

9/20/2011

1

Ceva’s Theorem

MA 341 – Topics in GeometryLecture 11

Ceva’s TheoremThe three lines containing the vertices A, B, and C of ABC and intersecting opposite sides at points L, M, and N, respectively, are concurrent if and only if

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Ceva’s Theorem

321-Sept-2011 MA 341 001

Page 2: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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2

Ceva’s Theorem

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Ceva’s Theorem

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Ceva’s Theorem

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Page 3: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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Ceva’s Theorem

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Ceva’s TheoremNow assume that

Let BM and AL intersect at P and construct CP intersecting AB at N’, N’ different from N.

821-Sept-2011 MA 341 001

Ceva’s TheoremThen AL, BM, and CN’ are concurrent and

From our hypothesis it follows that

So N and N’ must coincide.

921-Sept-2011 MA 341 001

Page 4: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

9/20/2011

4

M – midpoint AM=BM, N - midpoint BN=CNP - midpoint AP=CP

By Ceva’s Theorem they are concurrent.

Medians

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Theorem: In any triangle the three medians meet in a single point, called the centroid.

In ΔABC, let M, N, and P be midpoints of AB, BC, AC.Medians: CM, AN, BP

Orthocenter

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Let ΔABC be a triangle and let P, Q, and R be the feet of A, B, and C on the opposite sides.AP, BQ, and CR are the altitudes of ΔABC.

Theorem: The altitudes of a triangle ΔABC meet in a single point, called the orthocenter, H.

Orthocenter

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Page 5: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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Orthocenter

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By AA ΔBRC~ΔBPA (a right angle and B) BR/BP=BC/BAΔAQB~ΔARC (a right angle and A) AQ/AR=AB/ACΔCPA~ΔCQB (a right angle and C) CP/CQ=AC/BC

Orthocenter

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By Ceva’s Theorem, the altitudes meet at a single point.

Orthocenter

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Traditional route:BQ intersects AP.Now construct CH and let it intersect AB at R.Prove ΔARC~ΔAQBmaking R=90.

A

B

C

P

Q

R

H

Page 6: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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Incenter

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Let ΔABC be a triangle and let AP, BQ, and CR be the angle bisectors of A, B, and C.Angle Bisector Theorem: If AD is the angle bisector of A with D on BC, then

Incenter

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Proof: Want to use similarity.Where is similarity?

Construct line throughC parallel to AB

Incenter

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Proof: Want to use similarity.Where is similarity?

Construct line throughC parallel to AB

Extend AD to meet parallel linethrough C at point E.

Page 7: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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Incenter

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BAE CEA – Alt Int AnglesBDA CDE – vertical anglesΔBAD ~ ΔCDE – AATherefore

Note that CEA BAE CAE ΔACE isosceles CE = AC and

Incenter

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Let ΔABC be a triangle and let AP, BQ, and CR be the angle bisectors of A, B, and C.

Theorem: The angle bisectors of a triangle ΔABC meet in a single point, called the incenter, I.

Incenter

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Proof: Angle bisector means:

By Ceva’s Theorem we need to find the product:

Page 8: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

9/20/2011

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Incenter

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Thus by Ceva’s Theorem the angle bisectors are concurrent.

Circumcenter & Perp Bisectors

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Does Ceva’s Theorem apply to perpendicular bisectors?

Circumcenter & Perp Bisectors

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How can we get Ceva’s Theorem to apply to perpendicular bisectors?

Page 9: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

9/20/2011

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Circumcenter & Perp Bisectors

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Draw in midsegments

EF||BC perpendicular bisector of BC is perpendicular to EF is an altitude of ΔDEF

Circumcenter & Perp Bisectors

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Perpendicular bisectors of AB, BC and AC are altitudes of ΔDEF.

Altitudes meet in a single point perpendicular bisectors are concurrent.

CircumcircleTheorem: There is exactly one circle through any three non-collinear points.

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The circle = the circumcircleThe center = the circumcenter, O.The radius = the circumradius, R.Theorem: The circumcenter is the point of intersection of the three perpendicular bisectors.

Page 10: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

9/20/2011

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Question

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Where do the perpendicular bisectors of the sides intersect the circumcircle?

Question

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Where do the perpendicular bisectors of the sides intersect the circumcircle?At one end is point of intersection of angle bisector with circumcircleThe other end is point of intersection of exterior angle bisector with circumcircle.

Extended Law of SinesTheorem: Given ΔABC with circumradius R, let a, b, and c denote the lengths of the sides opposite angles A, B, and C, respectively. Then

3021-Sept-2011 MA 341 001

Page 11: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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Proof

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Three cases:

Proof

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Case I: A < 90ºBP = diameterΔBCP right triangleBP = 2R sin P = a/2RA = P 2R = a/sin A

Proof

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Case II: A > 90ºBP = diameterΔBCP right triangleBP = 2R sin P = a/2RA = P 2R = a/sin A

Page 12: Ceva’s Theorem - msc.uky.edu 11...Thus by Ceva’s Theorem the angle bisectors are concurrent. Circumcenter & Perp Bisectors 21-Sept-2011 MA 341 001 23 Does Ceva’s Theorem apply

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Proof

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Case III: A = 90ºBP = a = diameterBP = 2R2R = a = a/sin A

Circumradius and AreaTheorem: Let R be the circumradius and K be the area of ΔABC and let a, b, and c denote the lengths of the sides as usual. Then 4KR=abc

3521-Sept-2011 MA 341 001

Proof

3621-Sept-2011 MA 341 001

K = ½ ab sin C2K = ab sin Cc/sin C = 2Rsin C = c/2R2K = abc/2R4KR = abc


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