# Length of chord in circle - Which property

In the figure AB=4 , BC=6 , AC=5 and AD=6 what is length of DE ? Ans=9

I know there must be some property here that would solve this problem instantly but I cant figure it out any suggestions ? Edit: Since the two triangles are similar there corresponding sides will be equal in ratio , however I am still getting the wrong answer

BA   CA   BC
4    5    6
AE   6    DE


$$AE = \frac{24}{5}$$ and $$DE = \frac{36}{5}$$

Where am I going wrong ?

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You're applying the wrong similarity transform (a rotation instead of the reflection that's the actual similarity transform) and so getting the wrong cross ratios. You have AC:BC::AD:DE, (5:6::6:?) but in fact it's AB:BC::AD:DE (4:6::6:?). – Steven Stadnicki Jul 17 '12 at 22:36
One way to see why the symmetry should be reflection and not rotation is to consider moving A 'north' towards the top of the circle; then AC obviously shrinks, but on the other triangle the edge that shrinks is AE, not AD - so the similarity must be between ABC and ADE (note orientation), not ABC and AED. – Steven Stadnicki Jul 17 '12 at 22:42
Thanks that helped – Rajeshwar Jul 17 '12 at 22:42

Hint: Use the inscribed angle theorem.

Edit: I guess it's time to expand this answer a bit. The inscribed angle theorem guarantees that the angles $\angle CBE$ and $\angle CDE$ are equal, as well as the angles $\angle BCD$ and $\angle BED$. Also, the angle $\angle CAB$ is obviously equal to the angle $\angle EAD$. Therefore the triangles $\triangle AED$ and $\triangle ACB$ are similar. This means that the ratios of the lengths of respective edges should be equal or:

$$\frac{|AE|}{|AC|}=\frac{|AD|}{|AB|}=\frac{|ED|}{|CB|} \; .$$

This means that the length of edge $ED$ must be $9$.

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I know C and E share the same arc. But i cant figure how they help in determining DE ? – Rajeshwar Jul 17 '12 at 8:55
The triangles $\triangle ACB$ and $\triangle AED$ are similar because of that. – Raskolnikov Jul 17 '12 at 8:57
How is DE=9 if the two triangles are congruent ? – Rajeshwar Jul 17 '12 at 9:10
I meant similar, not congruent. Sorry for that. – Raskolnikov Jul 17 '12 at 9:12
Do I have to calculate the angles in triangle ABC ? – Rajeshwar Jul 17 '12 at 9:18