Complex geometry for me

O

OJean

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May 15, 2022
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Hello,

Is it possible to find a solution to this problem ?

Thanks,
OJeanfinalllllllllllll.png
 
It seems to me that a great deal of information is missing. For example, are line segments AD and DC parts of line segment AC? The diagram makes it appear that way, but is that true. Do we know anything about triangle ABC? The diagram makes it appear that the triangle is isosceles, but is that true. Is line segment DE parallel to line segment AB? Are line segments AD and CD component parts of line segment AC? Is D the midpoint of AC?
 
It seems to me that a great deal of information is missing. For example, are line segments AD and DC parts of line segment AC?

Yes, straight line.

Do we know anything about triangle ABC? The diagram makes it appear that the triangle is isosceles, but is that true.

Exact, isoceles.

Is line segment DE parallel to line segment AB?

Yes, it is.

Are line segments AD and CD component parts of line segment AC?

Yes.

Is D the midpoint of AC?

Not necessariy.

Thanks,
OJean
 
I'm finding it to be impossible to find AD. We would need to know something like AC to get the height of DEGF.

-Dan
 
OJean said:
Yes, parallel and isoceles.

Thanks.
Click to expand...
Given that, I can answer your question: Yes, it is possible.

Now, if you really want the solution, not just a yes/no, you'll need to follow the guidelines! (See #2.)

What have you tried? Where are you stuck? And, since you didn't put this under trigonometry, how much trig do you know? (It doesn't take much, mostly algebra.)

I can tell you that I worked with just one half of the figure, and some right triangles.
 
Dr.Peterson said:
Given that, I can answer your question: Yes, it is possible.

Now, if you really want the solution, not just a yes/no, you'll need to follow the guidelines! (See #2.)

What have you tried? Where are you stuck?

I can tell you that I worked with just one half of the figure, and some right triangles.
Click to expand...
Beat me to it!

topsquark said:
...get the height of DEGF.
Click to expand...
This is key. Hint: imagine line AB moving up and down vertically. This will affect the lengths of FG and AB differently (one increases as the other decreases). Therefore, you need to use the given equation "FG = AB/5" to lock in the height of DEGF.
 
@Cubist

I still do not get it. Did you assume that DF is the same length as EG?

As I understand it, that is not a given.
 
JeffM said:
@Cubist

I still do not get it. Did you assume that DF is the same length as EG?

As I understand it, that is not a given.
Click to expand...
Since angles FDE and GED are equal we get FD = EG. (If you drop a height from F to DE (which is the same h from G to DE) then etc.)

-Dan
 
Cubist said:
This is key. Hint: imagine line AB moving up and down vertically. This will affect the lengths of FG and AB differently (one increases as the other decreases). Therefore, you need to use the given equation "FG = AB/5" to lock in the height of DEGF.
Click to expand...
The first thing I did was to construct an approximate figure, without assuming an isosceles triangle, to confirm that was necessary for the answer to be constant; in doing so, I used exactly this idea, moving FG up and down.
JeffM said:
@Cubist

I still do not get it. Did you assume that DF is the same length as EG?

As I understand it, that is not a given.
Click to expand...
Given the assumptions we've elicited, the entire figure is symmetrical.
 
I have an easier solution. There is no solution as FGDE is not correct. Maybe FGED but not FGDE
 
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