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As shown in the image for a plane geometric problem: enter image description here

Could we prove $\angle ACD=\angle ABD$ without using the notion of circle?

It could seem easy if we have the notion of circle. But if we have no the notion of circle?

user49413
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  • Using similar triangles, you can prove that the midpoint of the hypotenuse of a right triangle is equidistant to the three vertices, which will allow you to show that the angles are congruent. That doesn't explicitly require the idea of a circle but only just, and you may need to be concerned about implicit requirements from the standard geometric theorems. – Michael Biro Apr 03 '19 at 15:38

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From $C$ draw $CE$ such that$$\angle BCE=\angle CBA$$Hence$$BE=CE$$Then since$$\angle CAB+\angle CBA=\angle ACB=90^o$$complements$$\angle CAB=\angle ACE$$and$$CE=AE$$making$$BE=CE=AE$$Thus $E$ is the midpoint of $AB$, and by the same construction and argument, since $\triangle ABD$, like $\triangle ABC$, is any right triangle with hypotenuse $AB$, a line from $D$ making with $DB$ an angle equal to $\angle DBA$ passes through the midpoint of $AB$, making$$BE=DE=AE$$and hence$$\angle EAD=\angle EDA$$ equal angles without circle Now consider $\angle ABD$ and $\angle ACD$:

Since $\angle ADB=90^o$ $$\angle ABD=90^o-\angle BAD$$

And$$\angle ACD=180^o-(\angle CAD+\angle CDA)=180^o-(\angle CAE+\angle CDE+2\angle EAD)$$that is, because of isosceles triangles $CEA$ and $CED$ $$\angle\ ACD=180^o-\angle ACD-2\angle EAD$$ Adding $\angle ACD$ to both sides$$2\angle ACD=180^o-2\angle EAD$$and dividing by $2$ $$\angle ACD=90^o-\angle EAD=90^o-\angle BAD=\angle ABD$$

This argument appears not to rest on any truths about the circle, although the auxiliary construction requires constructing angles equal to given $\angle ABC$ and $\angle ABD$, which in Euclid's treatment (Elements I, 23) does require drawing circles. Indeed, except for the drawing and extension of straight lines, it seems for Euclid all construction requires drawing circles, and arguments based on those constructions must appeal at least to the defining property of the circle, that all its radii are equal.

Edward Porcella
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  • This prove is OK. The existence of $AE$ is guaranteed by angle congruence axiom. How it can be drawn has no importance for the proof. From equal angle to equal facing side can be proven by using triangles congruence axiom. So, the notion of circle is not used. This proof shows that it's possible to use the property implied by a notion before the definition the notion. – user49413 Apr 05 '19 at 08:52
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It all depends on how you define angle. If you could propose some alternate definition then that might be interesting. But, if you wish to use the standard definition of angle then you're stuck with circles.

Here is the standard definition of the radian measure of an angle.

Given a plane $P$, a point $O$ in that plane, and a pair of points $X,Y \ne O$ in that same plane, the radian measure of the angle $XOY$ is defined as follows. Let $C$ be the circle of radius $1$ in $P$ centered at $O$. Let $x$ be the point where the ray $OX$ cuts through $C$. Let $y$ be the point where the ray $OY$ cuts through $C$. The points $x,y$ cut the circle $C$ into two circular arcs $\alpha$ and $\beta$. The angle $XOY$ is the minimum of the lengths of $\alpha$ and $\beta$.

As you can see, in this definition circles are at the heart of the very definition of angle. So if you had no notion of circle, then you would not have this definition of angle.

Lee Mosher
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  • I would say that there's a difference between angle and angle measure. Euclid (and Hilbert) defined angle and angle congruence without referring to circles. – Michael Biro Apr 03 '19 at 15:43
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    Agreed, although it was not clear to me what the OP intended. I was hoping that my answer might elicit his/her intentions more clearly. Also, one is always faced with the issue that Euclid not only did not define congruence of angles, but he did not even formulate sufficient axioms for it. – Lee Mosher Apr 03 '19 at 15:53
  • The equality of angle is in sense of congruence, but not in sense of their value. So, $x$ and $2\pi - x$ are considered the same angle. The notion of circle is not necessary. – user49413 Apr 03 '19 at 16:13
  • Notice that I carefully stated the equality of angle measure to be logically equivalent to congruence of angle: "The angle $XOY$ is the minimum of the lengths of $\alpha$ and $\beta$", which in the notation of your comment is the minimum of the two numbers $x$ and $2\pi - x$. – Lee Mosher Apr 03 '19 at 19:48