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This is a question from a school textbook (Extended Math for IGCSE by David Reyner), so I'm guessing it shouldn't be too hard to solve for you guys. Anyway, here it is:

In a triangle ABC, a line is drawn parallel to BC to meet AB at D and AC at E. DC and BE meet at X. Prove that: (a) the triangles ADE and ABC are similar. (Done) (b) the triangle DXE and BXC are similar. (Done) (c) AD / AB = EX / XB

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    It is true that DXE and BXC are similar, but sloppily expressed, since D is not what corresponds to B. It really should be "DXE similar to CXB".2011-09-15

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ok from a) you get $\frac{AD}{AB}=\frac{AE}{EC}=\frac{DE}{BC}....~(1)$ as the triangles are similar the ratio of their sides will be in same proportion. Now, similarly for b) you get $\frac{EX}{BX}=\frac{DX}{CX}=\frac{DE}{BC}....~(2)$ both $(1)\text{ and }(2)$ has $\frac{DE}{CB}~$ common $~\therefore \frac{AD}{AB}=\frac{EX}{XB}\quad$ as they both are equal to $\frac{DE}{BC}~$ from $(1) \text{ and }(2)$.

Hope this helps.

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Hint: From b), find another ratio equal to EX / XB.

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    Amaan, do you understand that if two triangles are similar, then the ratios of corresponding sides are all equal? So (b) gives you two ratios that are equal to EX/XB, and one of those two ratios relates to (a) as well....2011-09-15
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Hint 2: You have two pairs of simliar triangles. Is there any corresponding pair of edges that are in both pairs of triangles?

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    Helped a lot. Thanks.2011-09-15