*Originally posted by howzzat*

** The statement of the problem needs correction. It should be:-
**

AD/DB * BE/EC * CF/FA = 1. Otherwise it would be wrong . Take the simple example of an equilateral triangle ABC and let D, E, F be the mid points of AB, BC, and ...[text shortened]... ut AD/DB * BE/EC * CF/CA = 1/2.

It can be proved like this. In the given problem , let O be the common point of intersection of AE, BF & CD. Let us , for convenience, put:-

/_AOD = /_COE = a (say); /_BOD = /_COF = b (say);

/_BOE = /_AOF = c (say);

Also let us say, AD=x1;DB=x2;BE=x3;EC=x4;CF=x5;FA=x6;

OA=r1;OB=r2; and OC=r3;

/_ADO = t1; /_BDO = pi - t1 = t1' (say);

/_BEO = t2; /_CEO = pi - t2 = t2' (say);

/_CFO = t3; /_AFO = pi - t3 = t3' (say).

Now by the sine proportionality theorem of the property of triangles,

we have:

x1 / sin a = r1 / sin t1.

Hence x1 = r1 sin a /sin t1;

Similarly x2 = r2 sin b /sin t1'=r2 sin b / sin t1;

x3 = r2 sin c /sin t2;

x4 = r3 sin a /sin t2'=r3 sin a /sin t2;

x5 = r3 sin b / sin t3;

x6 = r1 sin c /sin t3'=r1 sin c /sin t3.

Now from here we get x1 x3 x5 = x2 x4 x6 =

= r1 r2 r3 sin a sin b sin c /( sin t1 sin t2 sin t3).

Hence (x1 x3 x5)/(x2 x4 x6) = 1

i.e.( x1/x2)*(x3/x4)*(x5/x6) = 1. This proves the desired relation.