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

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.

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).

• Simple and elegant. Thanks
– 1b3b
Oct 26, 2020 at 17:48
• @cosmo5 Is it not just RHS? Oct 26, 2020 at 17:49
• Wow, my bad! Been ages since I had a look at congruences. Oct 26, 2020 at 17:51
• SSA works if the angle is opposite to the larger side. In particular if it is a hypotenuse and the triangle is right.
– user700480
Oct 26, 2020 at 17:55

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)

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

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.

• Thanks, but I prefer synthetic geometry proof... Sorry, I forget to mention in the description. But keep the answer, it looks interesting!
– 1b3b
Oct 26, 2020 at 17:30
• 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... Oct 26, 2020 at 17:33

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.

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