Barukh.wrote:

> I would like to know the synthetic proof of the following statement:

>

> Given an arbitrary triangle ABC,

>

> let D be the foot of the perpendicular from A to BC

> let E be the foot of the perpendicular from D to AC

> let F be a point on the line DE

>

> Prove that AF is perpendicular to BE iff BD/DC = FE/FD.

Triangles EAD and EDC are similar. Consider the similarity that takes the

first into the second. It is the composition of 90 degrees rotation about E

and dilation with ratio

EC/ED = ED/EA (= tan CAD).

So it takes segment AF into the segment DH perpendicular to AF with H on EC

and EH/HC=EF/FD.

It remains to notice that EH/HC=BD/DC iff DH is parallel to BE, ie AF is

perpendicular to BE..

Regards,

Vladimir Dubrovsky