# Math Olympiad Geometry Question: Similar Triangles

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

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 $AE = \sqrt{AC^2 + CE^2}=25$.