Main content

### Course: High school geometry > Unit 4

Lesson 5: Solving problems with similar & congruent triangles# Challenging similarity problem

Interesting similarity problem where we don't have a lot of information to work with. Created by Sal Khan.

## Want to join the conversation?

- I am reeeally struggling with the exercises on 'Solving problems with similar and congruent triangles'. I feel like I'm missing some algebra knowledge or something, I can't even understand the Hints properly. Especially the exercise with Michelangelo's Doni Madonna pentagram, for example, how do you go from CD= x - x/phi to CD = (phi-1/phi)x??

Is anyone else having problems with these exercises, or are there some areas/videos I should refresh myself on? I've gone though this playlist, but I'm still struggling :/(21 votes)- Hi Gabrielle! (Beautiful name, I have a dear friend named Gabriella!)

I was having quite some difficulty with that too, because I wasn't sure how to write the proportions correctly even though I understand how to do the math behind it, so I was pretty stuck! If you could detail what your confusion was about with the hints and alike, I'd be happy to help you.

http://www.math10.com/en/geometry/similar-triangles/similar-triangles-theory-and-problems.html

This link helped me a lot (especially the practice at the end, which, I noticed, was esthetically a take-off of the question-answer-hints button from Khan Academy, all the way down to the smiley face when you get it right! Same colors, everything!)

I hope this helps you!

Your fellow learner,

@EuAmoAprender(17 votes)

- I ask if high school students are expected to solve the problem because I tried to and got stuck seemingly because I did not have the knowledge to apply algebra in this way. I suspect many are like me.

What can I do about this lack of ability to solve such problems that are challenging and new to me?(18 votes)- You should learn some algebra and some geometry more basic than this. The key is to to be able to put labels on things(6 votes)

- I understood the whole problem except the last thing he said at9:44, when he found the answer of 36/7, what did he mean when he said "regardless of how far and wide it is, it will always be 36/7"? Does this mean that for every problem like above the answer will always be 36/7, even if their angles or vertical length change? I find this hard to believe, and if somebody could explain it, I would be indebted to them.(7 votes)
- i recommend you get some squared paper and draw this problem to scale to prove it to yourself. make the left side 9cm and the right 12cm and use a ruler to connect the lines as in sal's diagram. now repeat but move both the lines closer to the middle , keeping them 9cm and 12cm. you'll notice when you connect the lines they pass through the same point. the length of y is unimportant. now repeat the diagram but make one side 4cm. you'll notice that the new CF has definately changed. the conclusions are: if the sides remain the same length you can move them as far apart or as close and CF will be the same. change the lengths and you'll get a new length CF(12 votes)

- Are high school students expected to solve this problem?(12 votes)
- I'm confused for when the number 1 suddenly pops into the problem when it says cf/9=1-x/y(8 votes)
- couldn't you also have done CF/9+CF/12=[(y-x)/y]+(x/y)? Then just simplifying that would get you the same answer.(7 votes)
- I see many comments in the section that say, just use 1/AB + 1/DE = 1/CF.

(I tried it, it works).

I've asked all of them why it works but haven't gotten an answer for months.

So could someone answer it here and help me understand the reason?

I would really appreciate it since this has been bugging me for a long time.(6 votes)- ok here it goes.

assume instead of 9 it was x

and instead of 12 it was y

i am following diagram for proof

see triangles ABC and DEC

ACE=DCE

AB//DE

SO CAB=DEC AND ABC=CDE

THESE TWO TRIANGLES ARE SIMILAR

WE CAN WRITE X/Y=CB/CD

now you can prove triangles BCF and DEB are similar easily

so CF/DE=CB/DB

we can CB and DC as xh and yh(h is a proportionality

constant)

now CF/y=xh/(xh+yh)

Now you can rewrite it as

1/CF=1/x+1/y(1 vote)

- I've never heard the letter Y so much in my entire life! O_O

Could someone find a more concrete example, I didn't understand what exactly he was trying to find or the method he was using.

Your answer will be very appreciated! :)(4 votes) - In this diagram, isn't triangle ABE similar to triangle DEB?(1 vote)
- They are actually not similar. Here is how I know:

Similar triangles have sides that are proportional. If the triangles were similar, then:`AB/DE = BE/EB`

We know that BE and EB are the same segment, so:

BE/EB = 1

AB/DE = 9/12

Since these two ratios are not the same, the triangles are not similar.(6 votes)

- I encountered a strange problem;

Since there are two right angles between ab and ed, they are parallel. That means that angle bac is congruent to angle dec. Angles acb and dce are congruent because they are vertical angles. From this, we can conclude that angle cbf and angle cef are congruent, because abc and dec are congruent and angles abf and def are right angles. That means that triangles dbf and cfe are congruent, by aas. But this also gives the congruence of acb and dce, also by aas. But we know this is not true, as we have corresponding sides are unequal. Is there a fault in my logic? If not, is this situation impossible? What can be adjusted to make it possible(while keeping the side lengths)?(3 votes)- > From this, we can conclude that angle cbf

> and angle cef are congruent, because

> abc and dec are congruent and angles

> abf and def are right angles

From the premises you presented prior to this, you cannot draw the conclusion that △ABC and △DEC are congruent. You have angle - angle - unequal-corresponding-sides, which isn't sufficient to satisfy any congruence postulate but does serve to prove similarity.

If instead you were talking about ∠ABC and ∠DEC that would still not be correct. ∠ABC is congruent to ∠EDC by alternating interior angles given that line BD is a transversal of parallel lines AB and DE.(1 vote)

## Video transcript

So given this diagram,
we need to figure out what the length of CF
right over here is. And you might already
guess that this will have to do something
with similar triangles, because at least it looks that
triangle CFE is similar to ABE. And the intuition there is it's
kind of embedded inside of it, and we're going to
prove that to ourselves. And it also looks
like triangle CFB is going to be similar
to triangle DEB, but once again, we're going
to prove that to ourselves. And then maybe we can
deal with all the ratios of the different sides
to CF right over here and then actually figure
out what CF is going to be. So first, let's prove to
ourselves that these definitely are similar triangles. So you have this
90-degree angle in ABE, and then you have this
90-degree angle in CFE. If we can prove just one
other angle or one other set of corresponding angles
is congruent in both, then we've proved
that they're similar. And there we can either show
that, look, they both share this angle right over here. Angle CEF is the
same as angle AEB. So we've shown two corresponding
angles in these triangles. This is an angle
in both triangles. They are congruent,
so the triangles are going to be similar. You can also show that this
line is parallel to this line, because obviously these
two angles are the same. And so these angles
will also be the same. So they're definitely
similar triangles, so let's just write that down. Get that out of the way. We know that triangle ABE
is similar to triangle CFE. And you want to make sure you
get it in the right order. F is where the
90-degree angle is. B is where the
90-degree angle is, and then E is where
this orange angle is. So CFE. It's similar to triangle CFE. Now let's see if we can figure
out that same statement going the other way, looking
at triangle DEB. So once again, you have
a 90-degree angle here. If this is 90, then
this is definitely going to be 90, as well. You have a 90-degree
angle here at CFB. You have a 90-degree
angle at DEF or DEB, however you want to call it. So they have one set
of corresponding angles that are congruent,
and then you'll also see that they both share
this angle right over here on the smaller triangle. So I'm now looking at this
triangle right over here, as opposed to the
one on the right. So they both share this
angle right over here. Angle DBE is the
same as angle CBF. So I've shown you
already that we have this angle is
congruent to this angle, and we have this angle
is a part of both. So it's obviously
congruent to itself. So we have two
corresponding angles that are congruent
to each other. So we know that this
larger triangle over here is similar to this smaller
triangle over there. So let me write this down. Scroll over to the
right a little bit. We also know that triangle DEB
is similar to triangle CFB. Now, what can we do from here? Well, we know that the ratios
of corresponding sides, for each of those
similar triangles, they're going to
have to be the same. But we only have one side
of one of the triangles. So in the case of ABE and CFE,
we've only been given one side. In the case of
DEB and CFB, we've only been given one
side right over here, so there doesn't seem to
be a lot to work with. And this is why this is a
slightly more challenging problem here. Let's just go ahead and see if
we can assume one of the sides, and actually,
maybe a side that's shared by both of
these larger triangles. And then maybe things
will work out from there. So let's just assume that
this length right over here-- let's just assume
that BE is equal to y. So let me just write this down. This whole length is
going to be equal to y, because this at least gives
us something to work with. And y is shared by both ABE
and DEB, so that seems useful. And then we're going to have
to think about the smaller triangles right over there. So maybe we'll call BF x. Let's call BF x. And then let's
FE-- well, if this is x, then this is y minus x. So we've introduced a
bunch of variables here. But maybe with all the
proportionalities and things maybe, just maybe,
things will work out. Or at least we'll have
a little bit more sense of where we can go with
this actual problem. But now we can start dealing
with the similar triangles. For example, so we want
to figure out what CF is. We now know that for these
two triangles right here, the ratio of the corresponding
sides are going to be constant. So for example, the
ratio between CF and 9, their corresponding
sides, has got to be equal to the
ratio between y minus x-- that's that
side right there-- and the corresponding side
of the larger triangle. Well, the corresponding
side of the larger triangle is this entire length. And that entire length
right over there is y. So it's equal to
y minus x over y. So we could simplify
this a little bit. Well, I'll hold
off for a second. Let's see if we can
do something similar with this thing on the right. So once again, we have CF,
its corresponding side on DEB. So now we're looking at
the triangle CFB, not looking at triangle CFE anymore. So now when we're
looking at this triangle, CF corresponds to DE. So we have CF over DE is going
to be equal to x-- let me do that in a different color. I'm using all my colors. It's going to be equal to
x over this entire base right over here, so this entire
BE, which once again, we know is y. So over y. And now this looks
interesting, because it looks like we have
three unknowns. Sorry, we know
what DE is already. This is 12. I could have written CF over 12. The ratio between
CF and 12 is going to be the ratio between x and y. So we have three unknowns
and only two equations, so it seems hard
to solve at first, because there's one
unknown, another unknown, another unknown, another
unknown, another unknown, and another unknown. But it looks like I can
write this right here, this expression, in
terms of x over y, and then we could
do a substitution. So that's why this
was a little tricky. So this one right here-- let me
do it in that same green color. We can rewrite it as CF over 9
is equal to y minus x over y. It's the same thing as
y over y minus x over y, or 1 minus x over y. All I did is, I essentially,
I guess you could say, distributed the 1 over y
times both of these terms. So y over y minus x over
y, or 1 minus x minus y. And this is useful, because
we already know what x over y is equal to. We already know that x over
y is equal to CF over 12. So this right over here, I can
replace with this, CF over 12. So then we get-- this is the
home stretch here-- CF, which is what we care about, CF over 9
is equal to 1 minus CF over 12. And now we have one
equation with one unknown, and we should be able to
solve this right over here. So we could add CF
over 12 to both sides, so you have CF over 9 plus
CF over 12 is equal to 1. We just have to find a
common denominator here, and I think 36
will do the trick. So 9 times 4 is 36, so if you
have to multiply 9 times 4, you have to multiply CF times 4. So you have 4CF. 4CF over 36 is the same
thing as CF over 9, and then plus CF over 12 is
the same thing as 3CF over 36. And this is going
to be equal to 1. And then we are left with
4CF plus 3CF is 7CF over 36 is equal to 1. And then to solve for
CF, we can multiply both sides times the
reciprocal of 7 over 36. So 36 over 7. Multiply both sides
times that, 36 over 7. This side, things cancel out. And we are left with--
we get our drum roll now. So all of this
stuff cancels out. CF is equal to 1 times 36
over 7, or just 36 over 7. And this was a
pretty cool problem, because what it shows you
is if you have two things-- let's say this thing is some
type of a pole or a stick or maybe the wall of a building,
or who knows what it is-- if this is 9 feet tall or 9
yards tall or 9 meters tall, and this over here,
this other one, is 12 meters tall or 12 yards or
whatever units you want to use, if you were to drape a string
from either of them to the base of the other, from the top
of one of them to the base of the other-- regardless of how far
apart these two things are going to be-- we just
decided they're y apart. Regardless of how
far apart they are, the place where those two
strings would intersect are going to be 36/7
high, or 5 and 1/7 high, regardless of
how far they are. So I think that was a
pretty cool problem.