Triangles

the key is the angle ACB=CEF.
That makes only one solution possible so I was wrong in my
earlier post. Don't even want to look at the other solutions given,
I am working my way through the maze of angles.

so it looks like ACE is isosceles,
AC is 14, CE=14, AB=25, CB=23, E bisects AB so AE is 12 1/2
and EB is also 12 1/2

Originally posted by ilywrin
hmm... parallel lines& an angle
Using Thales Theorem:
EF/AD=BF/BD=BE/AB = 8/15; Hence BD=15;
All that remains is CD=?
But I can't seem to plug in the angle in the equation 🙁
EDIT: One more thing though SIMILAR TRIANGLES
Namely if we ...[text shortened]... e derived from that is a question I am currently unable to answer.

Hey ilywrin amd others!

Isn't Thales' Theorem this:

An inscribed angle in a semicircle is a right angle.

So what has his theorem got to do with the ratio? Maybe wrong theorem? The ratio can be worked out because of similar triangles, as triangle EBF = triangle ABD. Thus DF = 7. That should get you going.

EDIT: Just read the rest of the posts and yes you are correct ilywrin BC = 23m. Well done.

Originally posted by Shanshu311
That would be the Pythagorean Theorum.

😉

Ummmm, where's the right angle????

Originally posted by elopawn
Hey ilywrin amd others!

Isn't Thales' Theorem this:

An inscribed angle in a semicircle is a right angle.

So what has his theorem got to do with the ratio? Maybe wrong theorem? The ratio can be worked out because of similar triangles, as triangle EBF = triangle ABD. Thus DF = 7. That should get you going.

EDIT: Just read the rest of the posts and yes you are correct ilywrin BC = 23m. Well done.

Thats what I thought, Thales figures are up this year!
I saw one version of Thales that talked about parallel lines and
triangles however. Here is the link:
http://www.math.washington.edu/~king/coursedir/m444a02/class/
10-21-thales.html
I think thats correct. I'll post it and then run the link myself and
see if it goes. I found it googling in Thales.
All in all a very nice problem. keep 'em coming!

the link works. I think this is the version of thales ilywrin used.

Actually what I had in mind was this Thales Theorem (good thing it was not Cauchy, then even google cannot help us sort thins out) which can be found here:
http://descartes.cnice.mecd.es/ingles/3rd_year_secondary_educ/Similarity/Semejan1.htm

Though I believe I cannot obtain the ratios using it 🙁 A common lapsus of memory and mixed up methods 🙂

Originally posted by ilywrin
Actually what I had in mind was this Thales Theorem (good thing it was not Cauchy, then even google cannot help us sort thins out) which can be found here:
http://descartes.cnice.mecd.es/ingles/3rd_year_secondary_educ/Similarity/Semejan1.htm

Though I believe I cannot obtain the ratios using it 🙁 A common lapsus of memory and mixed up methods 🙂

I may have entered something wrong but I couldn't get the link
to work. Would you verify this is correct?

I figured it out, I left in the www part. the wink lurks, I mean the
link wurks:'😉 Its a neat one, interactive but it requires a java
download at least for me. I have heard bad things about java
downloads as potentially having little hitchhicker trojans and such
so like when you make love, use a condom!

the gist of the descartes site on thales is if you have two parallel
lines with two lines not parallel cutting through the parallel lines,
if you converge one end of the cutting lines to a point, you now have
a triangle with similar ratios.

Originally posted by sonhouse
the gist of the descartes site on thales is if you have two parallel
lines with two lines not parallel cutting through the parallel lines,
if you converge one end of the cutting lines to a point, you now have
a triangle with similar ratios.

nicely summed up, nice to see that the problem generated a lot of interest, another's coming as I write....