# Prove similarity in a triangle

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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If a) and b) are done, then you don't need to see the solutions for these, right? Your only question is c) ? – J. M. Sep 15 '11 at 12:03
Yes, both questions are just given because they usually provide hints. – Some Guy Sep 15 '11 at 12:16
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". – Henning Makholm Sep 15 '11 at 12:50

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. – Some Guy Sep 15 '11 at 12:59

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

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I give up. I tried, but really, I suck at these proof sums. Any chance you could explain? – Some Guy Sep 15 '11 at 12:31
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.... – Gerry Myerson Sep 15 '11 at 12:44

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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