# Triangle split in half angle-wise

• Nov 3rd 2007, 04:26 AM
Optiminimal
Triangle split in half angle-wise
http://img528.imageshack.us/img528/8...rianglezx3.jpg

I'm searching for a simple expression for the distance CD in relation to BE (=2BD) in the picture above.

1) D is midpoint on BE. BE and hence BD = DE are known.

2) Angles BAC = CAE = BAE/2, that is, C represents the "angular midpoint" on BE as "seen from A". Those angles are known.

3) F is the center of a circle. Points A, B and E lie on the perimeter of that circle so AF = BF is its radius.

4) Angles BFD = BAE are known. Hence triangle BDF is completely known (because of right angle and BD are also known).

Number 4 above is actually necessary because the circle describes all positions of point A which yield constant angle BAE. (See PlanetMath: circumferential angle is half the corresponding central angle) I would be very interested in knowing if there exists a similar simple relationship for all points A such that instead CD remains constant!

The problem has implications for perspective. If you stand at A and look at BE, then BC will appear to be as long as CE. And I want to know what other points of observation A exist, which would remove the actual center D from the apparent center C with an equal distance CD. Alternatively, at least how CD relates to BD.

I hope that you find the question a little interesting!
:)
• Nov 3rd 2007, 06:03 AM
earboth
Quote:

Originally Posted by Optiminimal
http://img528.imageshack.us/img528/8...rianglezx3.jpg

I'm searching for a simple expression for the distance CD in relation to BE (=2BD) in the picture above.
1) D is midpoint on BE. BE and hence BD = DE are known.
2) Angles BAC = CAE = BAE/2, that is, C represents the "angular midpoint" on BE as "seen from A". Those angles are known.
3) F is the center of a circle. Points A, B and E lie on the perimeter of that circle so AF = BF is its radius.
4) Angles BFD = BAE are known. Hence triangle BDF is completely known (because of right angle and BD are also known).
Number 4 above is actually necessary because the circle describes all positions of point A which yield constant angle BAE. (See PlanetMath: circumferential angle is half the corresponding central angle) I would be very interested in knowing if there exists a similar simple relationship for all points A such that instead CD remains constant!

...

Hello,

I'm afraid but I believe there isn't a simple expression for the distance CD in relation to BE :mad:

For a given circle, a given distance EB (and that means a constant angle \$\displaystyle \angle(EAB)\$) the angle bisector of \$\displaystyle \angle(EAB)\$ passes allways through the point X. X is the intersection of EB and the perpendicular bisector of EB, that means point X is independent from the position of point A.
• Nov 3rd 2007, 07:13 AM
Optiminimal
Ah, okey. Thanks for the clear answer!
I will take a look at what solution I can find and see what I can make out of it.

Btw, what software are you using to make images like that? I just use Sketchup which isn't really made for geometrical purposes.
• Nov 3rd 2007, 07:37 AM
earboth
Quote:

Originally Posted by Optiminimal
...
Btw, what software are you using to make images like that?...

Hi,

have a look here: euklid dynageo homepage