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enter image description here

I actually made one however with the help of an ellipse.

Can the construction be done without using the concept of ellipse? I want another solution since this chapter problem in a book has not yet introduced the concept of ellipse so there maybe a solution.

To anyone asking how I did it with an ellipse here's how: construct a circle with center and end point on mid point and end point of the hypotenuse, respectively. This should act as the median to the hypotenuse as it is half of the hypotenuse (theorem), and the circle act as the locus of the third vertex. Now construct ellipse whose constant lenght is the sum of the base pivoted at the end points of the hypotenuse. The intersection of the circle and ellipse is the vertex that satisfy the condition. The angle between leg should be right by Thales theorem. So there you go, that is my construction.

enter image description here

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  • $\begingroup$ Also, I would like to add that construction should be done without assigning number values to the length. Theorem based approached is what I am after. $\endgroup$ – tighten May 11 '17 at 10:03
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Given: $a + b $ and $c$; angle $\gamma$ is $90°$ (opposite $c$) enter image description here

  1. Construct Thales circle over $c$, the basis of the triangle. $A$ is left endpoint of $c$, $B$ is right end point.

  2. Construct a circle about $A$ with radius = $(a+b)$.

  3. Construct the perpendicular bisector of $AB$. Let this line meet the Thales circle at $M$. Construct a circle about $M$ with radius = length $AM$ (= length $BM$).

  4. This circle about $M$, radius = length $AM$ intersects the circle about $A$ with radius = $(a+b)$ at $X$.

    Join $A$ and $X$ to get $AX$.

  5. Line $AX$ intersects Thales circle at $C$, the vertex of the right triangle with sides, $c, a $ and $b$.

Reasoning:

  1. Chord $AB$ subtends right angle at $M$, $M$ is on the Thales circle.

  2. Angle $AXB = 45°$, half the angle subtended at centre M.

  3. Consider triangle $BXC$. Angle $XCB =90°$ , supplementary angle. Hence angle $CBX = 45°$, triangle $XCB$ is isosceles, hence : length $CX$ = length $CB$.

Does this make sense?

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  • $\begingroup$ This is it! It took me a lot of time to understand your 2nd reasoning (haven't arrived at chapters regarding circle). Your proof is beautiful and I loved it. Now, I have to answer the next problem; this time it's the difference of the sides that is given instead. $\endgroup$ – tighten May 11 '17 at 14:45
  • $\begingroup$ MarianD.Thank you so much!!!How on earth did you dig this answer up after such a long time. What a nice drawing.Greetings, Peter $\endgroup$ – Peter Szilas Feb 23 '19 at 6:42
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Suppose the legs sum to $s$, and the hypotenuse is $h$. Then the side $x$ satisfies $x^2+(s-x)^2=h^2$, that's $2x^2-2sx+s^2-h^2=0$, a quadratic equation for $x$. Now you can use ruler and compass to construct sums, difference, products, quotients, and square roots, so you can construct $x$, and $s-x$, and from there it's easy.

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Here's my algorithm. Given $a+b$ and $\sqrt (a^2+b^2)$ construct $\sqrt(ab)$ and from that $ab$ then construct $(a+b)^2$ and $(a-b)^2$ and hence $a+b$ $a-b$. Then construct $a$ and $b$

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