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1) It's said that Thales proved that "a diameter divides a circle in two congruent parts". I searched a lot but did not find his proof. How can it be proved?

2) How to prove that "a circle is a convex figure"?

These are things that can be easily noted by drawing a circle and looking at it... But is it possible to prove them? For example, I found in a book that one can use the triangle inequality to show that the diameter is the largest chord... Everyone knows that, but I found it very interesting that there was a proof for that! :) Is there any book that discusses "proving the obvious"?

Thanks!

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  • $\begingroup$ For the first you may think about producing an isometry between $\{x^2+y^1\leqslant 1:ax+by< 0\}$ and $\{x^2+y^1\leqslant 1:ax+by< 0\}$. WLOG, you may assume $a=1,b=0$ upon a rotation. Thus, one can consider a reflection by the $x$-axis. For the second one you ought to use the fact that the Euclidean norm is indeed convex. $\endgroup$
    – Pedro
    Sep 28, 2013 at 22:34
  • $\begingroup$ What counts as a proof here? Do you allow coordinate geometry? $\endgroup$
    – Potato
    Sep 28, 2013 at 23:25
  • $\begingroup$ @Potato: I don't really know what counts as proof... I think that using other theorems to show that those two theorems are true. The simpler the theorems used, the best, IMO. Every answer is appreciated as when I don't know the mathematics used I search it on Google :D $\endgroup$
    – João Rimu
    Sep 28, 2013 at 23:50

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This is so basic that I’m not sure I can give a convincing answer. But the primitive notion of congruence between two Euclidean figures is that there should be a rigid motion of the plane (also allowing reflections) that transforms the first figure into the other. @Arash’s perfectly acceptable answer to your first question used a reflection, but let me give another answer.

Consider the $180^\circ$ rotation about the center of the circle. It transforms your specified diagonal into itself, since the diagonal is a line passing through the pivot-point center. And it transforms each point on the circumference of the one semicircle to a point on the circumference of the other semicircle. Similarly, points within the one semicircle get sent to interior points of the other. That should do it, and I think that Thales would recognize this as a proof.

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  • $\begingroup$ I think I understood! *Rotating a line segment 180º through its center maps it to itself. plus *EVERY diameter is mapped to itself though the 180º rotation though the center. So the two halves are congruent. I think I got it! $\endgroup$
    – João Rimu
    Sep 28, 2013 at 23:40
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Here we try to proving them, by using presentation of the point $(x,y)$ as a complex number $x+iy$. The arguments can be adapted to any other form of presentation of the circle.

WLOG, assume that the circle is centered at origin with unit radius.

For the first part, again WLOG, assume than the diameter is the line $y=0$. Now there is bijection from set of points in upper half-circle to set of points in lower half-circle namely $f(w)=\overline{w}$ where $w$ is complex number, representing the points in plane.

For the second part, take $w=aw_1+bw_2$ with $a+b=1$, for two points $w_1,w_2$ in the circle. WLOG assume $\mid w_1\mid\leq \mid w_2\mid$. Then: $$ \mid w\mid^2=(aw_1+bw_2)(a\overline w_1+b\overline w_2)\leq (a\mid w_1\mid+b\mid w_2\mid)^2\leq \mid w_2\mid^2 \leq 1 $$ therefore $w$ lies inside the unit circle and hence it is convex.

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  • $\begingroup$ For the first part, though obvious, you need to add that the bijection preserves distance. $\endgroup$
    – Macavity
    Sep 29, 2013 at 2:45

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