| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hello,
I remember studying a theorem that went something like:
Consider a triangle ABC with angular bisector AD. Then prove that
AB/BD = AC/CD.
Is this correct? I have not been able to prove this.
I dont have a book on geometry handy to check this out.
Murali.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1173.1 | Here's my 2 cents worth ... | BEEZER::MEGARITY | I remember when Rock was young | Wed Jan 03 1990 06:19 | 17 |
Yes, I think you're correct here. The proof itself is pretty simple.
Just extend BA beyond A to a point E such that CE is parallel to AD. It's
not difficult to see that angle AEC = angle DAB and angle CAD = angle ACE.
But since angle DAB = angle CAD - because AD is a bisector - this
makes triangle AEC isosceles. Thus AE = AC.
Now if you consider triangle CEB, it not difficult to see that AB/AE = BD/DC,
because AD is parallel to CE. But since AE = AC, we can then say that
AB/AC = BD/DC
QED ?
Ian M
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| 1173.2 | That was good! | COOKIE::MURALI | Wed Jan 03 1990 14:05 | 8 | |
QED ?
Yes. Thanks.
Since you are good at geometry, how about trying for a proof for the
pretty theorem in 20.3?!
Murali.
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