In the diagram below, △ABC and △CDE are two right-angled triangles with AC = 24, CE =7 and ∠ ACB = ∠ CED. Find the length of the line segment AE.


The above is the diagram.

I came across this question in a Math Olympiad Competition. I am able to find out that △ABC and △CDE are similar triangles but after that, I am not sure what to do to solve the question. Can anyone help me with the solution? Thanks.

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    $\begingroup$ What can you say about $\angle ACE$? $\endgroup$ – Blue Jun 4 '14 at 7:30
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    $\begingroup$ It is equal to 90 degrees? $\endgroup$ – snivysteel Jun 4 '14 at 7:33
  • $\begingroup$ @snivysteel Yes. $\endgroup$ – Cookie Jun 4 '14 at 7:33
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    $\begingroup$ Now use Pythagorean theorem: $$7^2+24^2=(\bar{AE})^2$$ and solve for $\bar{AE}$. $\endgroup$ – Cookie Jun 4 '14 at 7:34
  • $\begingroup$ Oh, I did not realise this question was so simple. Thanks ! =D $\endgroup$ – snivysteel Jun 4 '14 at 7:36

You don't even have to bother with similarity here (yes they are similar, but it doesn't matter).

Let $\angle ACB = \angle CED = \theta$. That means that $\angle ECD = 90^{\circ} - \theta$ by the angle sum of $\triangle CDE$.

That means that $\triangle ACE$ is a right triangle allowing to to apply Pythagoras' Theorem to it. So $AE = \sqrt{AC^2 + CE^2}=25$.


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