#### Chandgo Nevisas

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- Thread starter Chandgo Nevisas
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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.

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It could be any length, so therefore CD length is unknownDid you draw the diagram on graph paper?

Do you realise that CD could be any length?

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Thank You for your response.similar trianglesdo 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.

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AED, not ADE, is similar to CEB --

A useful fact to realize is that when triangles are similar, not only are their

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

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Thanks again for your response

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 theirsidesproportional, but so are theiraltitudes. 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. Butratiosof 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

AD:BC=AC:BD=DC:CD

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?

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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

In the same way, DEF and CEF are

Look for a pair of right triangles whose angles

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

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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

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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.

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 todo the work, and discover what you are able to do when you try.

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AED and CBE are not congruent. They are not even similar. It is AED and

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

Now look at triangles ACD and ECF (

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So angle ACD will be congruent to angle ECD.requiredto 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 andCEBthat are similar (notcongruent). This is because angles AED and CEB (the angles at E) are congruent by vertical angles,andangles 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 (notEFC). Don't you see (at least) two pairs of congruent angles there? One pair, in fact, are the very same angle!

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

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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.

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

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Are you asking for aI can see how angles ACD and ECD are the same angles. I would like to knowif they are any theorems or postulates. The same goes forADC and ECF. And to concludewhat would be the proportions. Their are only two side lengths given.

What are you saying about

Here is the

We have AD:BC = 40:30 = 4:3, so using similar triangles AED~CEB, AE:EC = 4:3 also.

We have ADC~EFC, so that the desired length EF can be found if we know the proportion: AD:EF = AC:EC. But AC = AE+EC, so AC:EC = (AE+EC):EC = (4+3):3 = 7:3. Can you use that to finish?