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I've been learning about triangle centers, and I have been able to prove that most of them exist. However, this one that I've come across has stumped a bit, and I haven't been able to find any proofs or hints towards how to prove this specific point always exists. I believe the point is called the Spieker center. The problem prompt goes a little like this.

Given a triangle $ABC$, construct lines parallel to the interior angle bisectors of angles $A$, $B$, and $C$, such that they pass through the midpoints of sides $BC$, $CA$, and $AB$ respectively. Prove that these lines are concurrent.

Any ideas on how I should approach this proof?

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  • $\begingroup$ To confirm: This is indeed called the Spieker center. $\endgroup$
    – Blue
    Apr 16, 2021 at 22:45

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The concurrency is evident using the medial triangle property as stated in the answer by cosmo5. Here is wiki link that you can refer to for more on it (wiki).

Here is a more elementary proof simply using, i) the midpoint theorem, ii) alternate interior angles theorem which states that if two parallel lines are cut by a traversal, the alternate interior angles are congruent, and iii) Concurrency of angle bisectors theorem.

enter image description here

$D, E$ and $F$ are midpoints of the sides of $\triangle ABC$. Line through $AM$ is the angle bisector of $\angle A$. Line through $EN$ is parallel to $AM$ and through midpoint $E$ of side $BC$.

Using midpoint theorem, $ADEF$ is a parallelogram and $\angle DEF = \angle A$.

Using property of parallel lines, $\angle FEN = \angle FMA = \angle MAD = \frac{\angle A}{2}$. So line through $EN$ is angle bisector of $\angle DEF$ in $\triangle DEF$. By the same logic, parallel lines through midpoints to other two angle bisectors of $\triangle ABC$ must be angle bisectors of $\angle DFE$ and $\angle EDF$.

Hence the proof is complete by concurrency of angle bisectors of a triangle.

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Rotate the original triangle $ABC$ by $180^\circ$ about its centroid $G$ and scale down to half its size. It will coincide with the triangle formed by its midpoints $D,E,F$.

The angle bisectors of $ABC$ would go to angle bisectors of $DEF$. Since any line rotated by $180^\circ$, a half turn, remains parallel to itself, the angle bisectors of $DEF$ are the lines to be constructed in the question.

We know that angle bisectors of any triangle concur at a single point, namely its incenter. Hence the lines through midpoints and parallel to the angle bisectors of $ABC$ concur at the incenter of $DEF$. The incenter of $DEF$ is simply the image of incenter of $ABC$ under our transformation (a half turn about centroid followed by scaling by half). The proof is complete.

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  • $\begingroup$ +1. I got so caught-up in the bisector-parallels generalization that I looked right past the obvious "the Spieker center is the incenter of the medial triangle" approach for this specific case. :) ... Well, perhaps OP will still find the Extended Ceva Theorem helpful in exploring more circle centers. $\endgroup$
    – Blue
    Apr 17, 2021 at 6:01
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    $\begingroup$ @Blue But your answer and the links therein were quite useful. Thankyou for writing those previous answers! :) $\endgroup$
    – cosmo5
    Apr 17, 2021 at 6:03
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    $\begingroup$ It's worth noting that angle-bisector-ness, per se, isn't required for this argument. The rotate-and-scale transformation guarantees that any trio of cevians through $A$, $B$, $C$ transfers to correspondingly-arranged parallel cevians through $D$, $E$, $F$; and, thus, the first trio's point of concurrence (if it exists, which it does in the case of the angle bisectors) transfers to the second's. $\endgroup$
    – Blue
    Apr 17, 2021 at 9:15
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    $\begingroup$ That is very nice! So we have two-sided correspondence : parallel lines through $A,D \iff $ corresponding cevians. $\endgroup$
    – cosmo5
    Apr 17, 2021 at 10:28
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I'll exploit a result that I presented in this answer, with different notation:

enter image description here

Extended Ceva's Theorem. Defining ratios of signed lengths, $$\alpha^+ := \frac{|BA^+|}{|A^+C|} \qquad \beta^+ :=\frac{|CB^+|}{|B^+A|} \qquad \gamma^+ := \frac{|AC^+|}{|C^+B|}$$ $$\alpha^- := \frac{|CA^-|}{|A^-B|} \qquad \beta^- :=\frac{|AB^-|}{|B^-C|} \qquad \gamma^- := \frac{|BC^-|}{|C^-A|}$$ the lines $\overleftrightarrow{A^+B^-}$, $\overleftrightarrow{B^+C^-}$, $\overleftrightarrow{C^+A^-}$ concur iff $$\alpha^+\beta^+\gamma^+ + \alpha^-\beta^-\gamma^- + \alpha^+\alpha^- + \beta^+\beta^- + \gamma^+\gamma^- \;=\; 1\tag{$\star$} $$

If $A^+$, $B^+$, $C^+$ are midpoints, then $\alpha^+=\beta^+=\gamma^+=1$, and $(\star)$ reduces to $$\alpha^-\beta^-\gamma^- + \alpha^-+\beta^-+\gamma^- = 0$$ However, we'll find that the angle bisector context makes it natural to express the "$-$" ratios in terms of the "$+$" ones; so I won't invoke the midpoint property prematurely.


Focusing on the angle bisector from $A$ aspect, let $\overleftrightarrow{A^+B^-}$ be parallel to that bisector, and let it meet the extended side $\overline{AB}$ at $V$.

enter image description here

With $u$, $v$, $\theta$ as labeled in the figure, we can use the Law of Sines on $\triangle BA^+V$ and $\triangle CA^+B^-$ to write $$\frac{\alpha^+ u}{c+v} = \frac{\sin\frac12A}{\sin\theta} = \frac{u}{b-v} \quad\to\quad v = \frac{\alpha^+b-c}{\alpha^++1} \quad\to\quad \beta^-:=\frac{v}{b-v}=\frac{\alpha^+ b-c}{b+c}$$ Likewise, $$\gamma^-=\frac{\beta^+ c-a}{c+a} \qquad \alpha^-=\frac{\gamma^+ a-b}{a+b}$$

Substituting into $(\star)$, we can write $$\frac{(1 + \alpha^+)(1 - \beta^+ \gamma^+)}{a} + \frac{(1 + \beta^+ )(1 - \gamma^+\alpha^+)}{b} + \frac{(1 + \gamma^+)(1 - \alpha^+\beta^+ )}{c} = 0\tag{$\star\star$}$$ This general condition for concurrence of bisector-parallels through $A^+$, $B^+$, $C^+$ is clearly satisfied when $\alpha^+=\beta^+=\gamma^+=1$ (ie, when $A^+$, $B^+$, $C^+$ are midpoints). $\square$

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