Find Side AB: Calculate Using m & n Ratios

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The discussion revolves around calculating the ratio DE:AC in a triangle given the ratio CD:DB = m:n. Participants note that since DE is parallel to AC, the similarity of triangles can be used to establish relationships between the segments. However, confusion arises regarding the correct interpretation of the ratios and whether sufficient information is provided to solve the problem. One participant suggests that if AD:DB were given instead, the solution would be straightforward, indicating that the current problem may lack necessary details. Ultimately, the complexity of the problem is acknowledged, with participants expressing uncertainty about deriving a clear solution.
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Hello.
There is a problem:
"Through D on a side AB of the triangle ABC drawn a line parallel to AC intersecting BC in E. D is such that CD:DB = m:n. Find DE:AC"

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So it is easy to find out that DB:AB equals to DE:AC as DE and AC are parallel. Since DB = n, there is only one need to express AB in terms of n and m. But how to do it? I tried using similar triangles, but I can't get AB through it. The trapezoid ADEC also gives no results as the second diagonal is unknown.
 
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I think you need also ##AE## and the intersection of both diagonals ##AE## and ##CD## in, say ##F##. This will make things more complicated as there will be more lengths involved, but I don't see another way to get hold on ##CD##. Then use the intercept theorem.
 
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fresh_42 said:
I think you need also ##AE## and the intersection of both diagonals ##AE## and ##CD## in, say ##F##.
Thank you. Actually, I considered that there may be not enough information in the problem. Now I am convinced.
 
quee said:
Thank you. Actually, I considered that there may be not enough information in the problem. Now I am convinced.
I am not sure. It depends on how the result has to look like. E.g. I got for the quotient
$$\dfrac{DE}{AC} =\dfrac{n}{m} \cdot \dfrac{CD}{AD+DB}$$
but it's not clear whether this will do or not. I don't see a second equation for ##CD##.
 
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Since ##DE## is parallel to ##AC##, you can use the similarity of triangles ##ABC## and ##ADE##. You have from the proportion that is given:
$$AD:DB = m:n \Rightarrow nAD = mDB$$
This means that ##AB = AD + DB = \left(\frac{m}{n} + 1\right) DB##. So now you have the proportion between ##AB## and ##DB## and from similarity you get the proportion between ##DE## and ##AC##.
 
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Antarres said:
Since ##DE## is parallel to ##AC##, you can use the similarity of triangles ##ABC## and ##ADE##. You have from the proportion that is given:
$$AD:DB = m:n \Rightarrow nAD = mDB$$
This means that ##AB = AD + DB = \left(\frac{m}{n} + 1\right) DB##. So now you have the proportion between ##AB## and ##DB## and from similarity you get the proportion between ##DE## and ##AC##.
Did you misread the given information?

Statement says: ##CD:DB = m:n##, not ##AD:DB ##.
quee said:
D is such that CD:DB = m:n. Find DE:AC"
 
@SammyS Indeed, I completely misread it. Apologies to the OP. In that case the exercise seems more complicated.
Might be that the method fresh gave would work, I so far don't see a clear way to get the proportion.
 
Antarres said:
Indeed, I completely misread it. Apologies to the OP.

It is okay. Thank you for your reply.
If only $$AD:DB= m:n,$$ it would be completely obvious that $$\frac{DE}{AC} = \frac{n}{n+m}$$
But this is, unfortunately, a much harder problem. Or it may be an actual lack of some information in the problem.
 
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