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This question already has an answer here:

Q. Given two line segments of length a and b. Draw a line segment of length $\sqrt{ab}$ using a ruler and compass.

I didn't get any idea how to approach to the solution.

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marked as duplicate by Jaap Scherphuis, Lord Shark the Unknown, Trần Thúc Minh Trí, Javi, Théophile Apr 3 at 16:34

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Let the base of the large triangle be $a+b$, and the height $h$. By similarity of the small triangles,

$$\frac ha=\frac bh$$ so that $$h=\sqrt{ab}.$$

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Draw a line of length $a+b$. Construct the perpendicular line in the point they joint. The semicircle over $a+b$ cuts that perpendicular. Now the distance between that point and the joining point is $\sqrt{ab}$ due to Euclid.

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  • $\begingroup$ If you don't mind , Can you show it analytically please?? $\endgroup$ – user579689 Apr 2 at 11:16
  • $\begingroup$ See en.wikipedia.org/wiki/Geometric_mean_theorem, please. $\endgroup$ – Michael Hoppe Apr 2 at 11:29
  • $\begingroup$ The word "diameter" is maybe missing in "The semi-circle over" $\endgroup$ – Jean Marie Apr 2 at 13:53
  • $\begingroup$ Being not a native speaker of the English language: What is ambiguous in the phrase "semicircle over a line segment"? $\endgroup$ – Michael Hoppe Apr 2 at 13:56
  • $\begingroup$ Nothing ambiguous... Don't bother with my remark. $\endgroup$ – Jean Marie Apr 2 at 13:58
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Join segments of length a and length b together on the same line. Call where they join Point J. Call their Midpoint M. Construct a circle centered at M through either end point of a+b. Construct a line perpendicular to a+b through point J. Call where it intersects the circle point Q. The length of QJ is $\sqrt{ab}$ as proven elsewhere. These above constructions follow from Euclid's postulates I, III, SAS, and ASA. So the constructions should be valid in neutral geometry.

A related approach. Construct a rectangle having one sidelength a and one side length b. Find a square having the same area as the starting rectangle. The side length of thes square will be $\sqrt{ab}$.

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