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 ΔABC, angle B equals 20o. (The same triangle appears in a different configuration inside regular 18-gon.) Two lines AD and BE are drawn such that ∠CAD = 60o, whereas ∠ACE = 50o. Find ∠ADE. (Check a more extensive discussion of this problem and a relevant solution.)
From the diagram it is immediate that the answer is 30o.
In a similar vein consider another problem. In an isosceles ΔABC, ∠ABC = 80o. A point M is selected so that ∠MAC = 30o and ∠MCA = 10o. Find ∠BMC.
References
- D. Gale, Tracking the Automatic Ant, Springer-Verlag, 1998
- V. V. Prasolov, Essays On Numbers And Figures, AMS, 2000
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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(ΔAFK) = AF·AK·sin(∠BAD)/2, etc. which leads to
and the theorem follows.
The following solution was found by S. T. Thompson, Tacoma, Washington (see Honsberger, pp 16-18).
With the change of notations, in ΔA14OA15, two lines A15X and A14Y are drawn such that ∠A14A15X = 60o and ∠A15A14Y = 50o. The question in the above configuration is to determine ∠A15XY.
Draw a circle with center O and radius OA14. The chord A14A15 subtends a 20o 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:
- A14A2 passes through Y.
- 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 120o. Therefore, ∠OA14A2 = 30o. 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 = 20o. 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 = 70o. But then also ∠A14XA16 = 70o. On the other hand, ∠A14XA15 = 180o - ∠OA14A15 - ∠XA15A14. ∠A14XA15 = 180o - 80o - 60o = 40o. Finally, ∠A15XY = ∠A14XA16 - ∠A14XA15 = 70o - 40o = 30o.
References
- R. Honsberger, Mathematical Gems, II, MAA, 1976
Menelaus and Ceva
