## RE: [EMHL] T construction given

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• Dear Hyacinthists, Maybe this could help. But I wasn t able to follow it. Please notice that a simple translation to English is not enough. I wonder there
Message 1 of 3 , Aug 1, 2008
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Dear Hyacinthists,

Maybe this could help. But I wasn't able to follow it.

Please notice that a simple translation to English
is not enough. I wonder there isn't a typo
anywhere.

Best regards,
Luis

Lösung. Macht man mittelst eines Kreises um A
mit dem Radius = b auf der Verlängerung von c
die Linie AG = b, und ebenso AE = c, wo E in a
liegt, so ist das Hülfsdreieck BEG gegeben, da
BE = p-q und, wie sich leicht ergiebt, der
ist.

_______________________________

To: hyacinthos@yahoogroups.com
From: qed_texte@...
Date: Wed, 30 Jul 2008 13:03:32 +0000
Subject: [EMHL] T construction given

Dear Hyacinthists,

Here I am again with another TC.

In a triangle we have:

B---Ha---B'-----Ma------------C

where , and .
Then CB'=p-q. Now how to construct the
triangle?

Best regards,
Luis

_________________________________________________________________
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• Dear Hyacinthists, I repeat my text, because there are missed parts. Lösung. Macht man mittelst eines Kreises um A mit dem Radius = b auf der Verlängerung
Message 2 of 3 , Aug 1, 2008
• 0 Attachment
Dear Hyacinthists,

I repeat my text, because there are missed parts.

Lösung. Macht man mittelst eines Kreises um A
mit dem Radius = b auf der Verlängerung von c
die Linie AG = b, und ebenso AE = c, wo E in a
liegt, so ist das Hülfsdreieck BEG gegeben, da
BE = p-q und, wie sich leicht ergiebt, der
\measuredangle BEG = 1R + A/2 ist.

In a triangle we have:

B---Ha---B'-----Ma------------C

where AB=AB'=c, BHa=B'Ha, CHa=p, and BHa=q.
Then CB'=p-q.
Now how to construct the triangle?

Best regards,
Luis

_______________________________

To: hyacinthos@yahoogroups.com
From: qed_texte@...
Date: Fri, 1 Aug 2008 21:10:53 +0000
Subject: RE: [EMHL] T construction given

Dear Hyacinthists,

Maybe this could help. But I wasn't able to follow it.

Please notice that a simple translation to English
is not enough. I wonder there isn't a typo
anywhere.

Best regards,
Luis

Lösung. Macht man mittelst eines Kreises um A
mit dem Radius = b auf der Verlängerung von c
die Linie AG = b, und ebenso AE = c, wo E in a
liegt, so ist das Hülfsdreieck BEG gegeben, da
BE = p-q und, wie sich leicht ergiebt, der
ist.

_______________________________

To: hyacinthos@yahoogroups.com
From: qed_texte@...
Date: Wed, 30 Jul 2008 13:03:32 +0000
Subject: [EMHL] T construction given

Dear Hyacinthists,

Here I am again with another TC.

In a triangle we have:

B---Ha---B'-----Ma------------C

where , and .
Then CB'=p-q. Now how to construct the
triangle?

Best regards,
Luis

_________________________________________________________________
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