- 16 Jun '05 18:51

I may be all wet here but when I draw it out it looks like AD can*Originally posted by elopawn***In a triangle ABC, D and F are points on BC, and E is a point on AB such that AD is parallel to EF, AngleCEF = AngleACB, AD = 15m, EF = 8m and BF = 8m. Find the length of BC.**

be Perpendicular to EF but I don't see any way it could ever be

parallel. Those two lines cross inside the ABC triangle and doesn't

look to me like crossing lines can ever be parallel at least not in

OUR universe. - 16 Jun '05 19:17

looking at it again, could you have made a typo, meaning E is on*Originally posted by elopawn***In a triangle ABC, D and F are points on BC, and E is a point on AB such that AD is parallel to EF, AngleCEF = AngleACB, AD = 15m, EF = 8m and BF = 8m. Find the length of BC.**

a point AC instead of AB? that could allow the two lines to be

parallel but they would still be called AD and FE. - 16 Jun '05 19:38

Are you sure? I was able to draw it out, just fine.*Originally posted by sonhouse***I may be all wet here but when I draw it out it looks like AD can**

be Perpendicular to EF but I don't see any way it could ever be

parallel. Those two lines cross inside the ABC triangle and doesn't

look to me like crossing lines can ever be parallel at least not in

OUR universe.

- 16 Jun '05 19:56

Construct an isoceles triangle where the two equal length sides are each 8m long - have the third side be the bottom of the triangle (to make this easier). Label the bottom two vertices B & E, and the top vertice F. Draw a line parallel to EF and 15m long such that its two ends will be along the extended lines BF and BE. Label the ends of these two lines A (bottom) and D (top). Then extend BD out some as yet to be determined distance and where it ends will be point C. Connect C to A, and you have the problem as described above. Note that BE and BF are both 8m long, and BD and AD are both 15 m long. Now the hard part is when you draw the line from C to E, the angle ACB has to equal the angle CEF.*Originally posted by sonhouse***I may be all wet here but when I draw it out it looks like AD can**

be Perpendicular to EF but I don't see any way it could ever be

parallel. Those two lines cross inside the ABC triangle and doesn't

look to me like crossing lines can ever be parallel at least not in

OUR universe.

I;m with Shanshu, my geometry is just a little to rusty to finalize the solution (although I'm wondering how he calculated the length of BE....) - 16 Jun '05 21:08

Isn't that interesting! I just automatically drew it ABC, an*Originally posted by jebry***Hi sonhouse!**

just switch your points B and C and make sure D is closer to C than F is and then you can draw it. AB is the base of the triangle and C is the apex. So, going clockwise around the triangle starting at the lower left corner you would have A, C, D, F, B and E.

equalateral triangle with A in lower left hand corner, B on top and

C on right hand tip. They are marked going clockwise.

But if you draw it where they are going counterclockwise it puts

EF and AD in parallel like the original description. Learn something

new every day. Didn't know that would make a differance.

Haven't even started solving the problem as of yet, just didn't get

past step 1 - 16 Jun '05 22:20 / 2 editshmm... 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 denote the intersection of CE and AD with H, we can easily see that the angle CHD = CEF (since AD||EF), and from the condition we get

the angle ACB = CHD, and therefore the triangle ADC and CHD are similar.

What good can be derived from that is a question I am currently unable to answer. - 16 Jun '05 22:30

In this problem you can prove AD=DB, 15 units, so DF=7 units*Originally posted by elopawn***In a triangle ABC, D and F are points on BC, and E is a point on AB such that AD is parallel to EF, AngleCEF = AngleACB, AD = 15m, EF = 8m and BF = 8m. Find the length of BC.**

(FB is 8 so 15 minus 8 =7)

So this proves only that the triangle ADB is isosceles.

So building ADB says nothing about ABC, since there are an infinite

number of possible lengths of AC that will satisfy the ADB requirement. - 16 Jun '05 22:55

And of course (stupid me) The Similar Triangles are three:*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 denote the intersection of CE and AD with H, we can easily see that the angle CHD = CEF (since AD||EF), and from the co ...[text shortened]... are similar.

What good can be derived from that is a question I am currently unable to answer.

ADC, CHD, and EFC

Okay ratios:

CD/AD=HD/CD (unknown CD and HD )

CF/AD=EF/CD may be rewritten as 7/15+CD/15 = 8/CD,

Reworking that we get (CD = x for convenience)

7x+x^2=120 or x^2 +7x - 120 = 0;

Therefore the psoitive root, which is luckily 8 is the solution.

Thus BC = 8+15 = 23 - 16 Jun '05 22:57I agree with sonhouse there. But maybe we have to pay attention to the angles too, nobody did that so far. So my question is...the angle ACB means if you only look at that triangle, take the angle at the point C right? and for angle CEF you take that triangle CEF only and then take the angle at the point E?