# Find the length of EF using similar triangles (portion of amusement-park ride)

#### Chandgo Nevisas

##### New member
A portion of an amusement park ride called the loop is shown. Find the length of EF (Hint: use similar triangles)

Could anyone solve this.

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#### Dr.Peterson

##### Elite Member
Use the hint: What similar triangles do you see in the figure? What ratios can you work out from that?

One method would involve, at an intermediate step, determining the ratio DF:FC.

Please read this summary of our posting guidelines, to see why I am only asking questions and giving a small hint.

#### Denis

##### Senior Member
Did you draw the diagram on graph paper?

Do you realise that CD could be any length?

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#### Chandgo Nevisas

##### New member
Did you draw the diagram on graph paper?

Do you realise that CD could be any length?
It could be any length, so therefore CD length is unknown

#### Chandgo Nevisas

##### New member
Use the hint: What similar triangles do you see in the figure? What ratios can you work out from that?

One method would involve, at an intermediate step, determining the ratio DF:FC.

Please read this summary of our posting guidelines, to see why I am only asking questions and giving a small hint.

I know that conclude that triangle DEF is similar to triangle FEC by SAS. Angle DFC is congruent to angle CFE, side ED/ side EC = side DF/FC. I can also say that triangle ADE is similar to triangle CEB by SAS as well. Angle AED is congruent to CEB by vertical angles. Side AE/ Side EC = side BE/ED.

#### Dr.Peterson

##### Elite Member
No, DEF and FEC are not similar. You must not have typed what you meant; DFC isn't even relevant, not to mention congruent to CFE.

AED, not ADE, is similar to CEB -- order matters. And that is the pair you need to use.

A useful fact to realize is that when triangles are similar, not only are their sides proportional, but so are their altitudes. This makes it easier to get to the goal, though you could also get to DF:FC via a second pair of similar triangles.

Note that it doesn't matter what CD is; the only actual lengths you know, or need to know, are vertical. But ratios of horizontal lengths matter.

#### Chandgo Nevisas

##### New member
No, DEF and FEC are not similar. You must not have typed what you meant; DFC isn't even relevant, not to mention congruent to CFE.

AED, not ADE, is similar to CEB -- order matters. And that is the pair you need to use.

A useful fact to realize is that when triangles are similar, not only are their sides proportional, but so are their altitudes. This makes it easier to get to the goal, though you could also get to DF:FC via a second pair of similar triangles.

Note that it doesn't matter what CD is; the only actual lengths you know, or need to know, are vertical. But ratios of horizontal lengths matter.

To confirm, you are claiming that triangle AED is similar to CEB. Based on this claim,

I have couple of scenarios for similar triangles, correct me if I'm wrong

1) Triangle ADC is similar to Triangle BCD
So in this case, it would be

2) Triangle DEF is similar to Triangle CEF

DF:FC=ED:EC=EF:EF

I would like to know which option is correct. I was then thinking how to achieve the desired goal which is to find the length of EF. Will I have to use the Pythagoras theorem or the law of cosine to find the length?

#### Dr.Peterson

##### Elite Member
You don't seem to be clear on why certain triangles are or are not similar.

Do you see why all the angles of AED are congruent to the corresponding angles of CEB? That is why they are similar. Now, what pairs of sides are proportional? Write those proportions.

Do you see that there is no reason to think that ADC is similar to BCD? Only the right angles are congruent.

In the same way, DEF and CEF are not similar.

Look for a pair of right triangles whose angles are the same.

There will be no need for the Pythagorean theorem or trigonometry. It's all similar triangles.

#### Denis

##### Senior Member
Go here (you'll become an expert!):

#### Chandgo Nevisas

##### New member
Go here (you'll become an expert!):

Thank you for your help. I have done other triangle similarity problem and have gotten the correct answers. I have used various theorems such as SAS, AAS, and SSS. I just needed help on this question. It would it really help my thought process if someone could solve and explain the steps. It would clear many doubts and help make a clearer image.

#### Dr.Peterson

##### Elite Member
First, do as I have already asked, and show why the triangles I mentioned are similar, and write out the proportions you find from that. Specifically, note either AE:CE or DE:BE.

Then consider triangles ACD and ECF. What can you say about them? What does this tell you about AD and EF?

The way to gain confidence and understanding is to do the work, and discover what you are able to do when you try.

#### Chandgo Nevisas

##### New member
First, do as I have already asked, and show why the triangles I mentioned are similar, and write out the proportions you find from that. Specifically, note either AE:CE or DE:BE.

Then consider triangles ACD and ECF. What can you say about them? What does this tell you about AD and EF?

The way to gain confidence and understanding is to do the work, and discover what you are able to do when you try.
I can say that triangle AED and CBE are congruent by vertical angles. It triangles are congruent they have to be similar as well. The proportions would be DE:BE , AE:CE, and AD:BC. I cannot figure out how triangle ACD and triangle EFC would be similar.

#### Dr.Peterson

##### Elite Member
First, you are still not naming triangles in corresponding order, so that you are not clearly stating the similarity you presumably intend. Possibly in your educational system you are not required to name the vertices in the same order, but it is a good idea anyway, since it clarifies what you mean and what it takes to prove it.

AED and CBE are not congruent. They are not even similar. It is AED and CEB that are similar (not congruent). This is because angles AED and CEB (the angles at E) are congruent by vertical angles, and angles ADE and CBE (the corresponding angles at D and B) are congruent as alternate interior angles, since AD and BC are parallel (both perpendicular to CD, by implication). The same is true of angles DAE and BCE.

To make use of that, note that you know AD:BC = 40:30.

Now look at triangles ACD and ECF (not EFC). Don't you see (at least) two pairs of congruent angles there? One pair, in fact, are the very same angle!

#### Chandgo Nevisas

##### New member
First, you are still not naming triangles in corresponding order, so that you are not clearly stating the similarity you presumably intend. Possibly in your educational system you are not required to name the vertices in the same order, but it is a good idea anyway, since it clarifies what you mean and what it takes to prove it.

AED and CBE are not congruent. They are not even similar. It is AED and CEB that are similar (not congruent). This is because angles AED and CEB (the angles at E) are congruent by vertical angles, and angles ADE and CBE (the corresponding angles at D and B) are congruent as alternate interior angles, since AD and BC are parallel (both perpendicular to CD, by implication). The same is true of angles DAE and BCE.

To make use of that, note that you know AD:BC = 40:30.

Now look at triangles ACD and ECF (not EFC). Don't you see (at least) two pairs of congruent angles there? One pair, in fact, are the very same angle!
So angle ACD will be congruent to angle ECD.

#### Dr.Peterson

##### Elite Member
Angles ACD and ECD are the same angle. With reference to the triangles where they matter, I'd call them ACD and ECF. Also, of course, angles ADC and EFC are both right angles; and angles DAC and FEC are corresponding angles on a transversal of parallel lines.

I presume you now see that those triangles are similar, and how to use that fact?

#### Chandgo Nevisas

##### New member
Angles ACD and ECD are the same angle. With reference to the triangles where they matter, I'd call them ACD and ECF. Also, of course, angles ADC and EFC are both right angles; and angles DAC and FEC are corresponding angles on a transversal of parallel lines.

I presume you now see that those triangles are similar, and how to use that fact?
I can see how angles ACD and ECD are the same angles. I would like to know if they are any theorems or postulates. The same goes for ADC and ECF. And to conclude what would be the proportions. Their are only two side lengths given.

#### Dr.Peterson

##### Elite Member
I can see how angles ACD and ECD are the same angles. I would like to know if they are any theorems or postulates. The same goes for ADC and ECF. And to conclude what would be the proportions. Their are only two side lengths given.
Are you asking for a theorem that states that an angle is congruent to itself, or that two angles formed from the same two rays are the same angle, or what? I'm not sure what you want.