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So problem is next: we have a Trapezoid($ABCD$) and we need to find $BC$(small base length). We know only $AD$ wich is big base length(its $10$cm). E point wich is on $AC$(diagonal). $S(abe):S(aed) = 1:2$

This is basically my puzzle... I know its not too complicated but for me i find it difficult so... The main point is finding the $x$ and everything known is on description

It would be great if somebody shows me the way of solving it.

Kenta S
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Zura
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  • Heya, first of all, it would be a good idea to say what you've done and what concepts you are working and familiar with so that the people don't get mad and be able to help you. – asdf Jul 05 '18 at 06:47
  • Okey got it. So i've thought that this two triangle could be Similar triangles somehow but i have not any clue how is it possible. The main problem is i haven't seen anything like that. Its my summer homework. I ve already solved 79 problem but thats something different. Oh ye interesting point === Its somehow connected to >>>> Area<<<<< – Zura Jul 05 '18 at 07:00
  • How can both $AB$ and $AD$ be bases when they have a common point? Also, is $ E$ arbitrary? – asdf Jul 05 '18 at 07:02
  • Ye i am sorry its BC and AD . Oh i wish i could upload photoe. E is just a point nothing specific – Zura Jul 05 '18 at 07:04

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This is more of a hint:

What can you say about the ratios of the heights from $D$ and $B$ to the diagonal $AC$ using the ratio of the areas?

If you denote $M=AC\bigcap BD$ then what can you san about the ratio $\frac{BM}{MD}$ using the ration of the height as before?

Finally what can you say about the ration $\frac{BC}{AD}$?

The big thing in this problem is to use Tales' theorem which is the same as using similar triangles, so try to fill in the gaps above.

asdf
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  • Okey thank you for hints i need to imagine picture good and understand fundamentally what you said(yoou know understanding the math with fereign language isnt pice of cacke xD) – Zura Jul 05 '18 at 07:34
  • H1/H2 = ½. BM/MD = ½ (this part is little foggy i know BMC is similar of AMD but cant connect heights clearly ) BC/AD = ½ and answer is >>>>BC = 5 <<<<< if you dont mind please clear this fog xD – Zura Jul 05 '18 at 07:53
  • Is it somehow connected cause ABM and CMD has equal AREA? – Zura Jul 05 '18 at 08:00
  • Yee yee i get it now thank you man Hints was really helpful to learn how to solve it i appreciate that you just dont throw the right answer. – Zura Jul 05 '18 at 08:05
  • $\triangle BMC$ and $\triangle DMA$ are similar and the ratio is $\frac{1}{2}$ since the heights are in this ratio. Glad I could help – asdf Jul 05 '18 at 08:13
  • Ye ye that was really helpful. Still i have to solve 90 problem and every next one is little harder or equally hard so i think i ll post here frequently xD – Zura Jul 05 '18 at 08:24
  • I am sorry i havnt reuputations so i cant rate your answer – Zura Jul 05 '18 at 08:34