Hello everyone! I'm getting stuck at this problem and I dont know how to solve it. I'm just curious about it. Any help will be greatly appreciated

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- November 23rd 2012, 08:48 PMchubakuenoGeometry problems
Hello everyone! I'm getting stuck at this problem and I dont know how to solve it. I'm just curious about it. Any help will be greatly appreciated

Attachment 25877 - November 24th 2012, 04:59 AMBobPRe: Geometry problems
Could you resubmit, including a diagram, with labels for the various points so that we can refer to lines and angles ?

- November 24th 2012, 06:31 AMchubakuenoRe: Geometry problems
thanks for your attention, here is the diagram

Attachment 25900 - November 24th 2012, 12:08 PMearbothRe: Geometry problems
1. You are dealing with 2 right triangles: and .

So the point A is located on a half circle (circle of Thales) over BD and point A is located on a half circle (circle of Thales) over CE.

2. You certainely have made an exact drawing(?). See attachment.

a) The point A is the intersection of the green circle and the circle of Thales over BD. The center of the green circle has the coordinates with the radius (why?)

b) Draw the line AC. Construct a right angle in A on AC. The leg of this right angle intersects the prolonged line BD in E.

3. I've done the construction in a coordinate system so you can read the length of x.

4. The line segment BD is divided harmonically by C (inner point) and E (outer point). According to the harmonic partition you'll get the proportion:

With your values:

Solve for x. - November 24th 2012, 05:00 PMchubakuenoRe: Geometry problems
No, it was just illustrative, but that's the way the problem was given to me. Thank you!

- November 25th 2012, 02:41 AMBobPRe: Geometry problems
My solution is rather more workmanlike !

Start with the right angled triangle ABD and let the side AB = a.

Then

so

Now let the angle ACD = and use the sine rule in each of the triangles ABC and ACD.

From ACD,

so

From ABC,

so

Equate the two expessions for simplify, and you find that

and from which it follows that AD =

Call the angle ABD then from the triangle ABD, and from that you can deduce that and

Now, finally, turn your attention to the triangle at the other end, ADE.

The angle AED will equal so making use of the sine rule again,

so

so after cancelling and then substituting for and

we finish up with