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## Re: [EMHL] Triangle Proof

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• Dear all, ... Let G be on AC with BE//DG, thus BD/CD = EG/EC by theorem of Thales. F is on the altitude DE of ADG, hence AF is perpendicular to BE or to DG iff
Message 1 of 2 , Jun 30, 2004
Dear all,

>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.
>
>Regards,
>Barukh.
>
Let G be on AC with BE//DG, thus BD/CD = EG/EC by theorem of Thales.

F is on the altitude DE of ADG, hence AF is perpendicular to BE or to DG
iff F is the orthocenter of ADG iff FG is perpendicular to AD, i.e. FG//DC.

By theorem of Thales FG//DC iff FE/FD = GE/GC.

Friendly,

Gilles Boutte
• ... 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
Message 2 of 2 , Feb 2, 2005
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,