# Geometry

Fiathahel
Posers and Puzzles 26 Aug '03 14:53
1. Fiathahel
Artist in Drawing
26 Aug '03 14:53
I've got a triangle ABC. On the line AB I take a point D, on BC I take E and on CA F. Show that if AE, BF, CD intersect in one point, AD/DB * BE/EC * CF/CA = 1.

It is easy to prove for some special cases, but I can't do it in general.
2. 10 Dec '04 19:13
Originally posted by Fiathahel
I've got a triangle ABC. On the line AB I take a point D, on BC I take E and on CA F. Show that if AE, BF, CD intersect in one point, AD/DB * BE/EC * CF/CA = 1.

It is easy to prove for some special cases, but I can't do it in general.
please give some hints...Will it invove the similarity of triangles?
3. TheMaster37
Kupikupopo!
11 Dec '04 18:03
Among other things, yes.

Fiathahel...isn't that in your geometry syllabus?
4. 22 Dec '04 07:45
Originally posted by Fiathahel
I've got a triangle ABC. On the line AB I take a point D, on BC I take E and on CA F. Show that if AE, BF, CD intersect in one point, AD/DB * BE/EC * CF/CA = 1.

It is easy to prove for some special cases, but I can't do it in general.
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 CA respectively. Obviously AE, BF and CD meet at the common point O which is the centroid of triangle ABC.
Now in this case obviously: AD/DB * BE/EC * CF/FA =1,
But AD/DB * BE/EC * CF/CA = 1/2.
5. 23 Dec '04 10:091 edit
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 &amp; 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.