Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 676.

## Wednesday, October 12, 2011

### Problem 676: Circles, Diameter, Tangent, Chord, Metric Relations

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ReplyDeleteConnect DB, DE and note that AB perpendicular to BC, DB per. To AC, DE per. To AB and EF per. To AC

We have DB^2=AD.DC= a.b ( relations in right triangle)

So DB=SQRT(a.b)

Triangle AEF similar to tri. ABD and triangle AED similar to tri. ABC

So we have x/DB=AE/AB=AD/AC=a/(a+b)

And x=a.DB/(a+b)=a.SQRT(a.b)/(a+b)

Peter Tran

Nice short solution!

ReplyDeleteI needed more work :-(

x /√(ab) = x / BD = AE / AB = AD / AC = a /(a+b)

ReplyDelete(∵ BD² = AD.DC; EF ∥ BD; DE ∥ CB)

We have CB _|_ to AB => CB^2 = CD.CA

ReplyDelete=> CB = Sqrt(b.(a+b))

and AB = Sqrt(a.(a+b))

Connect DE and consider the right triangle ADE. ADE is similar to ACB

=> AE/AB = AD/AC

=> AE = a.Sqrt(a.(a+b))/(a+b)

=> AE = a.Sqrt(a)/Sqrt(a+b)

Now consider the right triangle AFE which is similar to ABC

=> EF/CB = AE/AC

=> EF = x = [a.Sqrt(a)/Sqrt(a+b)].Sqrt(b.(a+b))/(a+b)

=> x = a.Sqrt(ab)/(a+b)