$△ABC$ is isosceles if bisector of $∠A$ bisects $BC$?

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

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 $SQC = p$

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

3. $QSC = \pi - 2p$

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

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

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

7. 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.

8. 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)