# Two Cevians and Proportions in a Triangle, II

Here is a problem #37 from the 1965 Annual High School Contest

Point D is selected on side AB of triangle ABC in such a way that

DF/CF + AF/EF is

(A) 4/5 (B) 5/4 (C) 3/2 (D) 2 (E) 5/2

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Copyright © 1996-2017 Alexander Bogomolny

Point D is selected on side AB of triangle ABC in such a way that

DF/CF + AF/EF is

(A) 4/5 (B) 5/4 (C) 3/2 (D) 2 (E) 5/2

### Solution

Draw the third cevian BK:

Then, by Ceva's theorem

AD/BD · BE/CE · CK/AK = 1

so that

1/3 · 2/1 · CK/AK = 1

from which CK/AK = 3/2.

CF/DF = CE/BE + CK/AK = 1/2 + 3/2 = 2 and

AF/EF = AK/CK + AD/BD = 2/3 + 1/3 = 1.

It follows that DF/CF + AF/EF = 1/2 + 1 = 3/2.

(Another solution is available eslewhere.)

### References

- C. T. Salkind,
*The Contest Problem Book II*, MAA, 1996

### Menelaus and Ceva

- The Menelaus Theorem
- Menelaus Theorem: proofs ugly and elegant - A. Einstein's view
- Ceva's Theorem
- Ceva in Circumscribed Quadrilateral
- Ceva's Theorem: A Matter of Appreciation
- Ceva and Menelaus Meet on the Roads
- Menelaus From Ceva
- Menelaus and Ceva Theorems
- Ceva and Menelaus Theorems for Angle Bisectors
- Ceva's Theorem: Proof Without Words
- Cevian Cradle
- Cevian Cradle II
- Cevian Nest
- Cevian Triangle
- An Application of Ceva's Theorem
- Trigonometric Form of Ceva's Theorem
- Two Proofs of Menelaus Theorem
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Menelaus from 3D
- Terquem's Theorem
- Cross Points in a Polygon
- Two Cevians and Proportions in a Triangle, II
- Concurrence Not from School Geometry
- Two Triangles Inscribed in a Conic - with Elementary Solution
- From One Collinearity to Another
- Concurrence in Right Triangle
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Copyright © 1996-2017 Alexander Bogomolny

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