Trigonometric Form of Ceva's Theorem

Ceva's theorem provides a unifying concept for several apparently unrelated results. The theorem states that, in ABC, three Cevians AD, BE, and CF are concurrent iff the following identity holds:

The theorem has a less known trigonometric form

sin(ABE)/sin(CBE) · sin(BCF)/sin(ACF) · sin(CAD)/sin(BAD) = 1,

or

sin(ABE) · sin(BCF) · sin(CAD) = sin(CBE) · sin(ACF) · sin(BAD).

The latter may serve as a source of great many trigonometric identities - some obvious, some much less so. Some unexpected identities can be obtained by placing the six points A, B, C at the vertices of a regular polygon. If the Cevians are extended to intersect the circumscribed circle, they become three diagonals (or sides) of the regular polygon. We are of course concerned with the case where a regular polygon has three concurrent diagonals.

For example, the original configuration in the applet below, suggests the following identity:

which can be easily verified. Play with the applet to find more such identities. The number of sides of the polygon, can be modified by clicking on the number - originally 18 - in the lower left corner of the applet.

The applet also lends itself to discovery of problems of a different kind. Return to the 18-gon. You may observe that the configuration now is reminiscent of a very popular problem. Namely, in an isosceles ABC, angle B equals 20^o. (The same triangle appears in a different configuration inside regular 18-gon.) Two lines AD and BE are drawn such that CAD = 60^o, whereas ACE = 50^o. Find ADE. (Check a more extensive discussion of this problem and a relevant solution.)

From the diagram it is immediate that the answer is 30^o.

In a similar vein consider another problem. In an isosceles ABC, ABC = 80^o. A point M is selected so that MAC = 30^o and MCA = 10^o. Find BMC.

References

1. D. Gale, Tracking the Automatic Ant, Springer-Verlag, 1998
2. V. V. Prasolov, Essays On Numbers And Figures, AMS, 2000

Copyright © 1996-2008 Alexander Bogomolny

Proof of Ceva's Trigonometric Identity

AF/FB · BD/DC · CE/EA = 1

is equivalent to

Area(AFK)/Area(BFK) · Area(BKD)/Area(CKD) · Area(CKE)/Area(AKE)

This is because triangles AFK and BFK have the same altitude from the vertex K and similarly for other two pairs of triangles. On the other hand, Area(AFK) = AF·AK·sin(BAD)/2, etc. which leads to

sin(ABE)/sin(CBE) · sin(BCF)/sin(ACF) · sin(CAD)/sin(BAD) = FB/AF · DC/BD · EA/CE = 1,

and the theorem follows.

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem

In an isosceles ABC, ABC = 80^o. A point M is selected so that MAC = 30^o and MCA = 10^o. Find BMC.

Now, it's obvious that BMC = 70^o.

Copyright © 1996-2008 Alexander Bogomolny

The following solution was found by S. T. Thompson, Tacoma, Washington (see Honsberger, pp 16-18).

With the change of notations, in A₁₄OA₁₅, two lines A₁₅X and A₁₄Y are drawn such that A₁₄A₁₅X = 60^o and A₁₅A₁₄Y = 50^o. The question in the above configuration is to determine A₁₅XY.

Draw a circle with center O and radius OA₁₄. The chord A₁₄A₁₅ subtends a 20^o arc, so that A₁₄A₁₅ is a side of the regular 18-gon inscribed into that circle. I numbered the vertices of that 18-gon as shown in the diagram above.

Two observations are important for the proof:

1. A₁₄A₂ passes through Y.
2. A₁₀A₁₆ passes through both X and Y.

Indeed, A₁₀A₁₆ = A₁₄A₂, as chords subtending equal arcs. Furthermore, they are symmetric with respect to radius OA₁₅. Therefore, they intersect on that radius. In the isosceles triangle OA₁₄A₂, the angle at O is obviously equal 120^o. Therefore, OA₁₄A₂ = 30^o. We see that A₁₄A₂ passes through Y.

Further, A₁₃ is the middle of the arc A₁₀A₁₆. Therefore, A₁₀A₁₆  OA₁₃.

Let's for the moment denote the point of intersection of A₁₀A₁₆ with OA₁₄ as X'. Since every point on A₁₀A₁₆ is equidistant from O and A₁₃, so is X': OX' = X'A₁₃. In the isosceles triangle OXA₁₃, OA₁₃X = A₁₃OX = 20^o. Therefore, X' = X, which proves the second of the two observations.

Now, as we've seen, in the isosceles OXA₁₃, A₁₀A₁₆ is a height to side OA₁₃. It's then bisects OXA₁₃, which implies OXA₁₀ = 70^o. But then also A₁₄XA₁₆ = 70^o. On the other hand, A₁₄XA₁₅ = 180^o - OA₁₄A₁₅ - XA₁₅A₁₄. A₁₄XA₁₅ = 180^o - 80^o - 60^o = 40^o. Finally, A₁₅XY = A₁₄XA₁₆ - A₁₄XA₁₅ = 70^o - 40^o = 30^o.

References

1. R. Honsberger, Mathematical Gems, II, MAA, 1976

Copyright © 1996-2008 Alexander Bogomolny