Find the ratio x:y. (lengths of sides in geometric figure)

[chijioke]

Please I need help with this. I was asked to find the ratio of x:y in the picture below
I then lettered the picture for easy referencing.

This is what did.
From triangle BDE, $$\frac{\text{|BF|}}{\text{|FE|}}=\frac{\text{|BC|}}{\text{|CD|}}$$$$\frac{4}{12}=\frac{6}{\text{t}}$$t = 18
From triangle EAB,
$$\frac{\text{|CE|}}{\text{|AC|}}=\frac{\text{|FE|}}{\text{|BF|}}$$$$\frac{15}{\text{u}}=\frac{12}{4}$$u = 5
The challenge is that I find the value x and y; I know that
$$\frac{15}{5}=\frac{12}{4}=\text{is} ~ 3:1$$What do I to get the ratio x:y?

 

[blamocur]

What can yo say about triangles ABC and CDE?

 

[chijioke]

I can see that they are similar triangles. But I am beginning to doubt it when I checked and the corresponding sides are not in the same ratio.
$$\frac{u}{t}=\frac{6}{15}=\frac{x}{y}$$$$\frac{5}{18}=\frac{\cancel{6}^2}{\cancel{15}_5}=\frac{x}{y}$$Can you see that? Are they really similar triangle as I thought?

 

[Dr.Peterson]

I can see that they are similar triangles. But I am beginning to doubt it when I checked and the corresponding sides are not in the same ratio.
$$\frac{u}{t}=\frac{6}{15}=\frac{x}{y}$$$$\frac{5}{18}=\frac{\cancel{6}^2}{\cancel{15}_5}=\frac{x}{y}$$Can you see that? Are they really similar triangle as I thought?

The triangles ABC and CDE are not similar, in the sense of corresponding vertices; nor is that true of ABC and DEC, as you appear to be assuming. It is ABC and EDC that are similar.

Do you see why angle A does not correspond to C or to D, but to E?

So, what ratios should be equal?

 

[khansaheb]

Please I need help with this. I was asked to find the ratio of x:y in the picture belowView attachment 37018
I then lettered the picture for easy referencing.
View attachment 37019
This is what did.
From triangle BDE, $$\frac{\text{|BF|}}{\text{|FE|}}=\frac{\text{|BC|}}{\text{|CD|}}$$$$\frac{4}{12}=\frac{6}{\text{t}}$$t = 18
From triangle EAB,
$$\frac{\text{|CE|}}{\text{|AC|}}=\frac{\text{|FE|}}{\text{|BF|}}$$$$\frac{15}{\text{u}}=\frac{12}{4}$$u = 5
The challenge is that I find the value x and y; I know that
$$\frac{15}{5}=\frac{12}{4}=\text{is} ~ 3:1$$What do I to get the ratio x:y?

Other than the given figure, any other information (e.g. AB|| DE & CF |E etc.) given?

 

[The Highlander]

View attachment 37019
This is what did.
From triangle BDE, $$\frac{\text{|BF|}}{\text{|FE|}}=\frac{\text{|BC|}}{\text{|CD|}}$$$$\frac{4}{12}=\frac{6}{\text{t}}$$t = 18
From triangle EAB,
$$\frac{\text{|CE|}}{\text{|AC|}}=\frac{\text{|FE|}}{\text{|BF|}}$$$$\frac{15}{\text{u}}=\frac{12}{4}$$u = 5
The challenge is that I find the value x and y; I know that
$$\frac{15}{5}=\frac{12}{4}=\text{is} ~ 3:1$$What do I to get the ratio x:y?

I have reworked below (in red) your original working. When I first looked at your OP, I thought: "This is all wrong! " but when I reworked it in what (to my mind is the correct way) I got the same answers for both u & t!
Examining you method more closely I realized how it would work (to find u & t) but, I have to say, it's a most unusual approach and it appeared to lead you to some erroneous conclusions further down. Please compare how I have dealt with getting u & t (comparing sides) and how that differs from your method (comparing side segments). I would suggest you adopt my method for future work. (But there's more than one way to skin a cat, of course, so feel free to continue merrily on with your own method if you prefer; just be aware of the pitfalls inherent in it. )





This is what did I would have done.
From triangle ΔBDE ~ ΔBCF, therefore... $$\frac{\text{|BF|}}{\text\color {red}{|BE|}}=\frac{\text{|BC|}}{\text\color {red}{|BD|}}$$$$\frac{4}{\color {red}16}=\frac{6}{\text\color {red}{|BD|}}$$$\displaystyle \color {red}\text{|BD| = 24 ⇒ t = 24 - 6 ⇒}$ t = 18

From triangle ΔEAB ~ ΔECF, therefore...
$$\frac{\text{|CE|}}{\text\color {red}{|AE|}}=\frac{\text{|FE|}}{\text\color {red}{|BE|}}$$$$\frac{15}{\text\color {red}{|AE|}}=\frac{12}{\color {red}16}$$$\displaystyle \color {red}\text{|AE| = 20 ⇒ u = 20 - 15 ⇒}$ u = 5

The challenge is that I find the value x and y; I know that
$$\frac{15}{5}=\frac{12}{4}=\text{is} ~ 3:1$$No! That is the ratio of side segments to side segments in ΔEAB! (A corollary of your unusual approach. )
If you had done it my way you would have been saying...$$\frac{20}{16}=\frac{15}{12}=\frac{5}{4}\implies\text{ their Ratio is} ~ 5:4$$which is the ratio of the two similar triangles ΔEAB & ΔECF; ie: all the sides in ΔEAB are 1.25 times longer than the corresponding sides in ΔECF.

What do I to get the ratio x:y?

(I'm switching away from red and back to black now as this an extension to what has gone before and is solely my contribution. )

Well, to get that ratio you need to consider two different similar triangles from the ones you've been using so far...

I think it's been made clear already that the ones you need to consider now are ΔABC & ΔEDC. (Simply because those are the triangles that include x & y as sides.)

Now that you have established the lengths of u & t, you now know the lengths of two sides in each of those triangles so the only unknowns now are, in fact, x & y!

At this point I would like to elaborate further on what @Dr.Peterson said above.
I would happily name the biggest triangle in the diagram as: CDE or DEC or CED or DCE or EDC or ECD (because all of these combinations do identify that particular figure) but, when we are dealing with identifying two triangles as being similar, he makes a very good point: that it is preferable to name the two triangles in terms of their corresponding angles.

Thus we should say: ΔABC ~ ΔEDC rather than, for example, ΔABC ~ ΔECD. (That probably couldn't be marked wrong but is definitely imperfect!)

Here is what is meant by naming similar triangles in terms of their corresponding angles...
In the diagram above:-

$\displaystyle \angle A = \angle E\\ \angle B=\angle D \quad ~\implies \Delta ABC \sim \Delta EDC\\ \angle C=\angle C$

Note that reading downwards gives you the correct naming for each triangle to ensure that the letters for corresponding angles in each triangle's name 'match up', yes?

It would, of course, then be perfectly proper to say, instead, that: $\displaystyle \Delta CAB \sim \Delta CED$ or any other combination as long as the corresponding angles 'match up' in the names.

However, an important corollary of naming similar triangles this way is that, not only do the order of the letters in the names identify which angles are the same, ordering the names this way also identifies which sides correspond too! And that is what you are most interested in now (to find the ratio x:y).

Can you see that (when we say: $\displaystyle \Delta ABC \sim \Delta EDC$) then side AB corresponds to side ED (first 2 letters), side BC corresponds to side DC (last 2 letters) and side AC corresponds to side EC (first & last letters).

Does that help you now to figure out what the ratio x:y will be?

 

[Dr.Peterson]

Examining you method more closely I realized how it would work (to find u & t) but, I have to say, it's a most unusual approach and it appeared to lead you to some erroneous conclusions further down. Please compare how I have dealt with getting u & t (comparing sides) and how that differs from your method (comparing side segments). I would suggest you adopt my method for future work. (But there's more than one way to skin a cat, of course, so feel free to continue merrily on with your own method if you prefer; just be aware of the pitfalls inherent in it. )

Actually, @chijioke's method for getting t and u is perfectly good; he is not using similar triangles directly, but rather a theorem (perhaps taught more in his country than in yours or mine) called the Basic Proportionality Theorem or the Side-Splitter Theorem, that parallel lines split the sides of a triangle equally. In this case, using triangle DEB, it says that |BF|/|FE| = |BC|/|CD|, just as he said. In fact, this problem seems designed to use that theorem, based on which segments are given.

I try not to tell people to do things my way, when their method is based on something they know too well to set aside. As I sometimes tell students, "The best way to solve a problem is the way you see -- which is why I ask you to show me your own thinking, so I can help you use that rather than the thinking that fits my own mind."

The only error is in the last part, where the wrong correspondence was used. (Possibly, by focusing on the BPT, students are not taught well enough how to recognize similar triangles.) In my experience, teachers would in fact mark a similarity statement as wrong if the wrong correspondence is shown, because they teach that as part of the meaning of similarity; and I think there's good reason for that. (But I would not do so if I had reason to think they were taught otherwise, and they showed they did see the correspondence.)

Other than the given figure, any other information (e.g. AB|| DE & CF |E etc.) given?

The arrow do indicate those parallel lines; but I agree it's better if a problem explicitly states everything that is to be assumed, rather than trust the diagram to say everything; and something could have been missed by showing only the diagram.

 

[chijioke]

Actually, @chijioke's method for getting t and u is perfectly good; he is not using similar triangles directly, but rather a theorem (perhaps taught more in his country than in yours or mine) called the Basic Proportionality Theorem or the Side-Splitter Theorem, that parallel lines split the sides of a triangle equally. In this case, using triangle DEB, it says that |BF|/|FE| = |BC|/|CD|, just as he said. In fact, this problem seems designed to use that theorem, based on which segments are given.

Yes, just, like you noted, what I used to obtain the value of t = 18 and u = 5 was the theorem of proportional division of the sides of a triangle which states that a line drawn parallel to one side of a triangle divides the other two sides in the same ratio.
But I was not able to use the same theorem to obtain the value of x and y to enable me get the ratio x:y.
Coming to the matter of applying the concept of ∼ Δs, I must thank you The Highlander and Dr.Peterson for widening my horizon on the idea behind ∼ Δs as to what ∼ triangles are and what they are not.
Looking at the smaller Δ, Δ EAB which consists of two ∼ Δs ECF and EAB - guess I'm calling it well now. The ratio of the corresponding sides is $$\frac{\text{|EC}|}{\text{|EB|}}=\frac{\text{|EF|}}{\text{|EB|}}=\frac{15}{15+u}=\frac{12}{12+4}$$$$\rightarrow 12(15+u)=15\times 16 \\ u = 20-15=5$$From the bigger Δ, Δ BED, there are two ∼
Δs BFC and BED.
The ratio of the corresponding sides is
$$\frac{\text{|BF|}}{\text{|BE|}}=\frac{\text{|BC|}}{\text{|BD|}}=\frac{4}{16}=\frac{6}{6+t}$$$$\rightarrow 4(6+t)=16\times 6 \\ t = 24-6=18$$Now t= 18, and u =5, the same I got when I applied the theorem of proportional division of the sides of a Δ. The issue now is how to find the value of x and y just I stated earlier.

It is ABC and EDC that are similar.

I am still lost here. I still finding it difficult understanding how ΔABC and EDC are ∼Δs. I was thinking that there are only two pairs of ∼Δs which are (a) ΔECF and ΔEAB, (b)ΔBFC and ΔBED. The only way I can be convinced that ΔABC and ΔEDC is only by looking at the ratio
$$\frac{\text{|BC|}}{\text{|CD|}}=\frac{\cancel{6}^1}{\cancel{18}_3}=\frac{6}{t}=\frac{\text{|AC|}}{\text{|CE|}} \\=\frac{u}{15}=\frac{\cancel{5}^1}{\cancel{15}_3}=\frac{x}{t} \\ \rightarrow x:y =1:3$$ Waoh! it worked.
How can I look at these two Δs


and call them ∼ . Perhaps because they are flipped. If so this kind of pairs ∼ Δs are not easy to recognize by me. Please could you please paste some link where I can handle exercises with this type of pairs ∼ Δs?

@Dr.Peterson, you suggested that I should use the concept of ∼ Δs
Okay let me give it a try. Looking at the smaller Δ EAB, the ratio of the corresponding sides of the ∼ Δs it consists of is
$$\frac{\text{|EC|}}{\text{|EA|}}=\frac{\text{|EA|}}{\text{|EB|}}=\frac{15}{(15+u)}=\frac{12}{16} \\ =\frac{\cancel{15}^3}{\cancel{20}_4}=\frac{\cancel{12}^3}{\cancel{16}_4} \\ \rightarrow 3:4$$Because I want to get the value of x, I extend
$$\frac{15}{20}=\frac{12}{16}=\frac{\text{|CF|}}{\text{|AB|}} \\ =\frac{3}{4}=\frac{\text{|CF|}}{|x|} \rightarrow 3:4$$Problem! What is the value of |CF| and |X|? Would I be right to say that |CF|=3 and |x|=4?
Coming to the bigger Δ, ΔBED which consists of two similar Δs, ΔBFC and ΔBED, the ratio of the corresponding sides is
$$\frac{\text{|BF|}}{\text{|BE|}}=\frac{\text{BC}}{\text{|BD|}}= \\ \frac{4}{16}=\frac{6}{t+6}=\frac{4}{16}=\frac{6}{24} \\ =\frac{\cancel{4}^1}{\cancel{16}_4}=\frac{\cancel{6}^1}{\cancel{24}_4} \rightarrow 1:4$$I am interested in obtaining the value of y, I extend again
$$\frac{4}{16}=\frac{6}{24}=\frac{\text{|FC|}}{\text{|ED|}}=\frac{\text{|FC|}}{\text{|y|}} \rightarrow 1:4$$Problem again! What could be the value of |FC| and |y| respectively. Could be I right to say that |FC| =1 and |y| =4?
Oh lest I forget, ratio x:y is demanded not necessarily the values of x and y.
May be the ratio is 1:3. I am correct now to say that x:y =1:3?
Suppose I am correct, I would like to ask here, apart from applying the idea of |
∼ Δs, what other thing, idea or theorem could I have applied to get the ratio of x:y?
Thanks for helping in the first place.

 

[Dr.Peterson]

I am still lost here. I still finding it difficult understanding how ΔABC and EDC are ∼Δs. I was thinking that there are only two pairs of ∼Δs which are (a) ΔECF and ΔEAB, (b)ΔBFC and ΔBED. The only way I can be convinced that ΔABC and ΔEDC is only by looking at the ratio
$$\frac{\text{|BC|}}{\text{|CD|}}=\frac{\cancel{6}^1}{\cancel{18}_3}=\frac{6}{t}=\frac{\text{|AC|}}{\text{|CE|}} \\=\frac{u}{15}=\frac{\cancel{5}^1}{\cancel{15}_3}=\frac{x}{t} \\ \rightarrow x:y =1:3$$ Waoh! it worked.
How can I look at these two Δs
View attachment 37060View attachment 37062

and call them ∼ . Perhaps because they are flipped. If so this kind of pairs ∼ Δs are not easy to recognize by me. Please could you please paste some link where I can handle exercises with this type of pairs ∼ Δs?

To see why these triangles are similar with the correspondence we talked about, you need to look for pairs of congruent angles -- congruent because they are alternate interior angles between parallel lines:

​

When you see that, you can draw (or just imagine) them in corresponding orientation:



I don't have time now (it's too late) to examine your last response to me. But I don't see where you get |EC/|EA|=|EA|/|EB|. Please correct that, and perhaps everything will follow.

 

[The Highlander]

Yes, just, like you noted, what I used to obtain the value of t = 18 and u = 5 was the theorem of proportional division of the sides of a triangle which states that a line drawn parallel to one side of a triangle divides the other two sides in the same ratio.
But I was not able to use the same theorem to obtain the value of x and y to enable me get the ratio x:y.

That is precisely why I (and, I believe, @Dr.Peterson & @blamocur) were 'steering' you towards the use of Similarity to solve this problem.

If you have another look at how I reworked your OP (here in Post #6) you should see that my approach is based solely on the similarity of the triangles involved.

(Let me re-state, for the avoidance of any doubt. it was not my intention to say that your approach was wrong, just that (as you, yourself, have now recognized) it does not facilitate working towards the final answer required of you; the theorem you employed is not one we specifically teach here (at HS level) and so I didn't immediately recognize that was what you were doing but did later say that I could see how it worked.)

Having re-read what I wrote (here in Post #6) I think it would have been more precise (because there are actually three different angles extant at point B) if I had written this...

​

Here is what is meant by naming similar triangles in terms of their corresponding angles...​

Referring just to the red & yellow triangles in the diagram above:-​

​

$\displaystyle \angle A = \angle E\\ \angle B=\angle D \quad ~\implies \Delta ABC \sim \Delta EDC\\ \angle C=\angle C$​

​

Note that reading downwards gives you the correct naming for each triangle to ensure that the letters for corresponding angles in each triangle's name 'match up', yes?​

I trust that @Dr.Peterson's latest post (immediately above) has now convinced you that those red & yellow triangles are, indeed, similar.

Again, considering just those two triangles, the angles at A & E are the same because they form alternate angles created by a straight line crossing two parallel lines (ie: Z-Angles). The same is true of the angles at B & D and, of course, where two straight lines cross (at Point C) vertically opposite angles are the same size.

Therefore, since the red & yellow triangles share the same three angles, they must be similar because any two triangles that share the same three angles are, by definition, Similar Triangles.
And, furthermore, if two triangles are similar then their corresponding sides are all in the same ratio.

That is what I was nudging you towards when I said (here in Post #6):-

"However, an important corollary of naming similar triangles this way is that, not only do the order of the letters in the names identify which angles are the same, ordering the names this way also identifies which sides correspond too! And that is what you are most interested in now (to find the ratio x:y).​

​

Can you see that (when we say: $\displaystyle \Delta ABC \sim \Delta EDC$) then side AB corresponds to side ED (first 2 letters), side BC corresponds to side DC (last 2 letters) and side AC corresponds to side EC (first & last letters)."​

Following on from that last sentence, I had hoped that you would have then seen that the red, green & blue lines are corresponding sides as now shown in the diagram below...

​

and, therefore, that...

$$\frac{\text {|\color{2C82C9}AB\color {black}|}}{\text {|\color{2C82C9}ED\color {black}|}}\left(=\frac{x}{y}\right)=\frac{\text {|\color{00A885}BC\color {black}|}}{\text {|\color{00A885}DC\color {black}|}}\left(=\frac{6}{t}=\frac{6}{18}=\frac{1}{3}\right)=\frac{\text {|\color{E25041}AC\color {black}|}}{\text {|\color{E25041}EC\color {black}|}}\left(=\frac{u}{15}=\frac{5}{15}=\frac{1}{3}\right)$$
So, can you now see how that shows what the ratio of x:y is?

May be the ratio is 1:3. I am correct now to say that x:y =1:3?
Suppose I am correct, I would like to ask here, apart from applying the idea of |
∼ Δs, what other thing, idea or theorem could I have applied to get the ratio of x:y?
Thanks for helping in the first place.

I trust you are now reassured that it is perfectly correct to say that: "x:y = 1:3"
But I am at a loss to understand why on earth you would wish to discover what "other thing, idea or theorem" you might use to solve this problem when Similarity works so well!
Do you just not like Similarity?

 

[chijioke]

Do you just not like Similarity?

No, I just what learn different ways of skinning a cat.

 

[chijioke]

But I don't see where you get |EC/|EA|=|EA|/|EB|. Please correct that, and perhaps everything will follow.

Sorry, that's an error. I was actually making reference to

What I intended writing is $$\frac{\text{|EC|}}{\text{|EA|}}=\frac{\text{|EF|}}{\text{|BF|}}=\frac{12}{16}$$

 

[chijioke]

So, can you now see how that shows what the ratio of x:y is?

I trust you are now reassured that it is perfectly correct to say that: "x:y = 1:3"

Yes. Thank you.

 

[Dr.Peterson]

Sorry, that's an error. I was actually making reference to
View attachment 37093
What I intended writing is $$\frac{\text{|EC|}}{\text{|EA|}}=\frac{\text{|EF|}}{\text{|BF|}}=\frac{12}{16}$$

No, I think you must mean $$\frac{\text{|EC|}}{\text{|EA|}}=\frac{\text{|EF|}}{\text{|{\color{Red}EB}|}}=\frac{12}{16}$$.

But I think you do have the right idea.

Again, I don't insist on using these similar triangles rather than what you did in the first part; but using them habitually does better prepare you for using them in the final step, where they are needed.

 

[chijioke]

No, I think you must mean $$\frac{\text{|EC|}}{\text{|EA|}}=\frac{\text{|EF|}}{\text{|{\color{Red}EB}|}}=\frac{12}{16}$$.

Yes. Thank you.

 

[chijioke]

What application did you use in getting the triangles ABC and DCE coloured?

Please what application did you use in getting the segments |AB|, |AE|, |BD|, and |DE| coloured?

 

[The Highlander]

What application did you use in getting the triangles ABC and DCE coloured?

Hi @chijioke, skinned any cats lately?

I used the excellent (and free) program Paint.net for both of these diagrams.

The 'Paint Bucket' tool in the program filled the triangles nicely with a single click.
I used two of the 27 (default) 'colours' (I don't consider black a 'colour') that are built into the Forum's palette; that enables me to (easily) change the colour of text to match my diagrams (though the Forum's palette does allow you to change text to any colour you like if you know its Hex RGB values).

I have added the little colour chart I put together based on the Forum's palette's colours that I use most often (so I can use those colours in my diagrams) if that's of any interest to anyone.

Please what application did you use in getting the segments |AB|, |AE|, |BD|, and |DE| coloured?

As I said above, this was also done in Paint.net but I'm wondering if you were so interested because you perhaps thought I had somehow changed the colour of the lines in your original picture. Maybe you thought I had some program that could exchange one colour for another?

Paint.net can actually do that but not always with what one might consider 100% success (and it's a bit fiddly; not ideal for changing the colour of the likes of the lines in your diagram).

No, all I did was draw new (coloured) lines over the existing ones but made them slightly thicker than the original lines; looking closely you can spot the difference in the the thickness of the coloured lines compared to the remaining black lines.

Again I used my FMH Colour Chart (see below) to match up the colours of my lines to the text: "the red, green & blue lines" and the LaTex fractions that followed.

Hope that helps.

NB: The numbers in brackets are just the Cartesian coordinates of the ticked colours.​

 

[chijioke]

I will try it later. But can coreldraw be used as well?

 

[The Highlander]

I will try it later. But can coreldraw be used as well?

Dunno.

I've never used CorelDRAW but I imagine it will be able to do pretty much the same things as Paint.net but at a considerably higher cost (unless you have pirated version )!

Looking at their website, CorelDRAW may well be 'packed' with many more (and powerful) features than Paint.net but, unless you're willing to risk downloading a pirate version of it or paying hundreds of pounds for a legal version, I'm at a loss to think of anything that might be needed that Paint.net can't do for free!

 

[chijioke]

Did you download the paint.net you are using or are you using it online ?