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Use similar triangles to prove that $|AS|:|SD|=|BS|:|SE|=2:1$.

My try:

$|AE|=|EC|$, $|BD|=|DC|$ $\Rightarrow \triangle ABC \sim \triangle EDC$

$\Rightarrow \angle A=\angle E \Rightarrow AB \parallel ED$

$\Rightarrow \angle BAD=\angle ADE$, $\angle ABE=\angle DEB$

$\Rightarrow \triangle ABS\sim \triangle DES$

From here, I can't prove the ratios..

Could someone please help me? Thanks in advance.


1 Answer 1


You know that $|AB|=2|ED|$, because $\frac {|AB|}{|ED|}=\frac{|AC|}{|EC|}=2$. Now, the ratio between $\triangle ABS$ and $\triangle DES$ is $2$ too, and thus it follows that $$\frac{|BS|}{|ES|}=\frac{|AB|}{|DE|}=\frac{|AC|}{|EC|}=2$$

  • $\begingroup$ I can't see how to deduct that the ratio between $\triangle ABS$ and $\triangle DES$ is 2 too.. $\endgroup$
    – rae306
    Aug 24, 2014 at 13:26
  • $\begingroup$ Please check all $=$-signs on the last line carefully and tell me which one you don't understand. $\endgroup$
    – Ragnar
    Aug 24, 2014 at 13:28
  • $\begingroup$ I already understand, thank you so much! ;-) Bedankt! $\endgroup$
    – rae306
    Aug 24, 2014 at 13:30
  • 1
    $\begingroup$ Have a look here: finaletrainingutrecht.tk (My training site of the Dutch Mathematical Olympiad.) At the bottom is a link to a book with similar exersizes. ('opgavenboekje') $\endgroup$
    – Ragnar
    Aug 24, 2014 at 13:36
  • 1
    $\begingroup$ Bedankt! Ga ik zeker gebruiken ;-) $\endgroup$
    – rae306
    Aug 24, 2014 at 13:42

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