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triangles ABC, ACD and BCD are right triangles, E is the midpoint of segment AB.

If AB = 20cm, find CE.

I'm having a hard time understanding these relationships between the sides of right triangles. Help again

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You posted an analogous question earlier today: math.stackexchange.com/questions/153070/… Did that answer not make sense? –  math-visitor Jun 2 '12 at 23:43
    
It helped yes, but I tried to follow similar format and can't come up with the answer. –  Jorge Jun 2 '12 at 23:59
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In its current form, $D$ and the triangles $ACD$ and $BCD$ seem redundant, and we do not have enough information to solve the problem. –  TMM Jun 3 '12 at 0:21
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Is there some sort of convention about where the right angle is when you write triangles with their vertices? Or is this intentionally ambiguous? –  rschwieb Jun 3 '12 at 0:40
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1 Answer 1

As a partial answer (since the situation is almost certainly underdetermined as written), if $\angle C$ in $\triangle ABC$ is a right angle, then $\overline{AB}$ is a diameter of the circle that circumscribes $\triangle ABC$, so $E$ is the center of the circle and $CE=\frac{1}{2}AB=\frac{1}{2}\cdot20\text{ cm}=10\text{ cm}$.

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Just to make things explicit, if the right angle in ABC is at A or B then there is nothing to be said about CE except that it's strictly greater than 10cm. –  Gerry Myerson Jun 5 '12 at 4:54
    
@GerryMyerson: Out of curiosity, if the right angle is at A or B, given that triangles ACD and BCD share side CD, can they both be right triangles? I'm not seeing an easy way to make this happen in the plane, but I also haven't tried very hard yet. –  Isaac Jun 5 '12 at 4:57
    
Since those other two triangles seemed to be irrelevant, I ignored them. But I think you may have discovered that they actually are relevant, if indeed they force the right angle in ABC to be at C. Well done! –  Gerry Myerson Jun 5 '12 at 5:49
    
@GerryMyerson: oh, were that it were true—let A be the right angle in triangle ABC, construct the circle with diameter AC and construct the line perpendicular to BC through C. Call the intersection of this circle and line D (the intersection that isn't C). Now we have all the givens satisfied, but cannot compute CE, right? –  Isaac Jun 5 '12 at 5:57
    
Right you are. Problem's busted. –  Gerry Myerson Jun 5 '12 at 6:33
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