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## High school geometry

### 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 :/ • 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!
@EuAmoAprender
• 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? • I understood the whole problem except the last thing he said at , 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. • 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
• Are high school students expected to solve this problem? • 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. • 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. • 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'm confused for when the number 1 suddenly pops into the problem when it says cf/9=1-x/y • In this diagram, isn't triangle ABE similar to triangle DEB?
(1 vote) • 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)?  