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Draw triangle $A, B, C$ with usual notation.

If bisector $s$ of the angle $\alpha$ intersects side $a$ in $S$ and if $|AS| = |BS|$ ($S$ is a midpoint of $a$) then triangle $ABC$ is an isosceles triangle. This is just my hypothesis, although I am pretty sure it's correct, but I need a proof.

Of course, if $ABC$ is indeed isosceles we have (by symmetry) that $S$ is midpoint of $a$ and it's intersection of $s$ and $a.$ I am interested in reverse statement mentioned above. Thanks

Forgot to mention: emphasis is on synthetic-geometry proofs.

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4 Answers 4

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Drop the heights from $S$ to $CA$ and $CB$ to get points $A', B'$.

Then, $\triangle CSA'\cong\triangle CSB'$, as right triangles with another angle equal ($\angle SCA'=\angle SCB'$) and a common side.

This implies that $SA'\cong SB'$.

Now look at $\triangle SAA'$ and $\triangle SBB'$: as they are right triangles with congruent sides $SA'\cong SB'$ and $SA\cong SB$ (as per assumption), then those triangles are also congruent, which gives you $\angle A=\angle B$.

(Strictly speaking, you would have two cases, where either interior $\angle A$ is equal to interior $\angle B$, or interior $\angle A$ is equal to exterior $\angle B$, but the second option is obviously impossible because in the second case the sum of all angles of $\triangle ABC$ would be bigger than $180^\circ$.)

This then implies that $\triangle ABC$ is isosceles (two equal angles).

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  • $\begingroup$ Simple and elegant. Thanks $\endgroup$
    – 1b3b
    Oct 26, 2020 at 17:48
  • $\begingroup$ @cosmo5 Is it not just RHS? $\endgroup$
    – player3236
    Oct 26, 2020 at 17:49
  • $\begingroup$ Wow, my bad! Been ages since I had a look at congruences. $\endgroup$
    – cosmo5
    Oct 26, 2020 at 17:51
  • $\begingroup$ SSA works if the angle is opposite to the larger side. In particular if it is a hypotenuse and the triangle is right. $\endgroup$
    – user700480
    Oct 26, 2020 at 17:55
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Your conjecture is correct, and can be proved in coordinates pretty easily. (If by "side a" you mean the side opposite the angle $\alpha$, which is located at point $A$, then you should have "$|CS| = |BS|$", rather than "$|AS| = |BS|$", of course. I'm pretty sure that this is what you meant.)

Note that for your conjecture to be true, we need $\alpha \ne 0$, so that $AC$ and $AS$ are distinct lines.

Here's a sequence that constitutes a proof, but you'll have to do your own drawing.

Assume $AC > AB$ (I'm going to skip the |X| symbols when it's obvious I'm talking about lengths.). Put a point $Q$ on $AC$, with $A-Q-C$ and $AQ = AB$. If we say that $x = BS$, then $x = QS$, as well, by reflecting across line $AS$. Let $H$ be a point on AS with $A-S-H$.

Letting $\beta$ be the angle at $B$, and $p$ be the angle $BSA$, we have

  1. $p = ASQ$ (ASA congruence)

1.5 $\gamma = SQA = \beta$

1.6 $SQC = p$

  1. $HSC= p$ (vertical angles, defn of $p$)

  2. $QSC = \pi - 2p$

  3. $QSC + SCQ + CQS = \pi$

  4. $(\pi - 2p) + SCQ + p = \pi$ (using item 1)

  5. $SCQ - p = 0$, so $SCQ = p$.

  6. The line $AS$ meets $BC$ with angle $p$ at $S$; the line $AC$ meets $BC$ with angle $p$ and $C$, so $AS$ and $AC$ are parallel.

  7. Contradiction, for $AC$ and $AS$ meet at $A$. (also using the "we need" that's boldfaced above, and which shows that $AC$ and $AS$ are not identical)

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  • $\begingroup$ Yes, angle "under" $A$ is $\alpha$, I made an error... $\endgroup$
    – 1b3b
    Oct 26, 2020 at 17:43
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By sine law:

$$\frac {\sin \alpha}{AS} = \frac{\sin \angle CAB}{CS}, \ \frac {\sin \alpha}{BS} = \frac{\sin \angle CBA}{CS}$$

Since $AS = BS$, we have $\sin \angle CAB = \sin \angle CBA$.

Hence either $\angle CAB = \angle CBA$ or $\angle CAB + \angle CBA = 180^\circ$.

The latter cannot happen since that would imply $\alpha = 0^\circ$.

The former implies that the triangle is isosceles.

I think there should be a simpler solution without invoking trigonometry, though.

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  • $\begingroup$ Thanks, but I prefer synthetic geometry proof... Sorry, I forget to mention in the description. But keep the answer, it looks interesting! $\endgroup$
    – 1b3b
    Oct 26, 2020 at 17:30
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    $\begingroup$ I know, that's why I dislike this solution. I think I have discussed this with some of my students before, I just need to jog my memory... $\endgroup$
    – player3236
    Oct 26, 2020 at 17:33
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I assume you mean $A$ and $B$ are endpoints of side $a$ since you say $S$ is the midpoint of $a$, but then that would deviate from "usual notation" since the endpoints of side $a$ should be $B$ and $C$.

If $a$ is $AB$, you can use the Angle bisector theorem to prove that $AC$ and $BC$ are equal, and hence, the triangle is isosceles:

$$\frac{|AS|}{|BS|}=\frac{|AC|}{|BC|} \quad 1=\frac{|AC|}{|BC|} \quad |AC|=|BC|$$

If however, you are using usual notation and mean that $S$ is the midpoint of $a$ such that $|AS|=|BS|$ and $|BS|=|CS|$, then the proof can not only be derived using the Angle bisector theorem but also by setting angles equal to each other:

Let half the angle $\alpha$ be $\theta$. Since $\Delta ACS$ and $\Delta ABS$ are both isosceles, that means angles $b=c=\theta \longrightarrow \Delta ABC$ is isosceles.

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  • $\begingroup$ You are right. Thanks $\endgroup$
    – 1b3b
    Oct 26, 2020 at 17:36
  • $\begingroup$ Note that the angle bisector theorem needs trigonometry to prove. $\endgroup$
    – player3236
    Oct 26, 2020 at 17:42
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    $\begingroup$ No, you can make additional constructions to the triangle to prove the Angle bisector theorem without trigonometry: mathbitsnotebook.com/Geometry/Similarity/SMAngle.html $\endgroup$ Oct 26, 2020 at 17:45
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    $\begingroup$ Someone really should add this proof to the Wikipedia article. $\endgroup$
    – player3236
    Oct 26, 2020 at 17:50
  • $\begingroup$ In my book from 1st grade (long time ago!) there is the picture of Oliver Byrne's proof (1847.) $\endgroup$
    – 1b3b
    Oct 26, 2020 at 17:57

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