Property of the $2$-$3$-$4$ triangle and its extensions

Question

I'm trying to answer this question but I'm fully stuck. I've tried to draw the diagram but it still is a mystery to me. Here is the problem:

a) Prove that the $2$-$3$-$4$ triangle has the property that one of the triangles formed by the lines containing two of its sides and the angle bisector of one of its external angles is similar to the original triangle.

b) Find all triangles with no side length greater than $10$ with this property.

c) Prove that there exists a triangle with this property with longest side length $n$ iff $n$ is not divisible by the square of any prime.

Answer 1

In the diagram $\triangle ABC$ is a $2$-$3$-$4$ triangle, but think of it also as a generic $a$-$b$-$c$ triangle. The exterior angle bisector at $A$ meets the extension of $\overline{BC}$ at $D$; let $d:= |\overline{CD}|$.

The Exterior Angle Bisector Theorem states

$$\frac{|\overline{DB}|}{|\overline{DC}|} = \frac{|\overline{AB}|}{|\overline{AC}|} \quad\to\quad \frac{a+d}{d} = \frac{c}{b} \quad\to\quad d=\frac{ab}{c-b} \tag{1}$$

For the $2$-$3$-$4$ triangle, we see that $d = 6$, and that $|\overline{BD}| = 8 = 2|\overline{AB}|$: the lengths of the sides of $\triangle DBA$ that surround $\angle B$ are in proportion $2:1$. That's true of $\triangle ABC$, as well! Consequently, $\triangle ABC \sim \triangle DBA$. That establishes the first part of your question.

In a generic $a$-$b$-$c$ triangle, the similarity property holds when the key ratio is $c:a$. That is, $$\frac{a+d}{c} = \frac{c}{a} \tag{2}$$

To find what triangles satisfy $(1)$ and $(2)$ simultaneously, we simply eliminate $d$:

$$a^2 = c(c-b) \tag{$\star$}$$

(Sanity check: $2^2 = 4(4-3)$.)

With that, I'll leave you to answer the remaining parts of your question.

Answer 2

Hint:

For part a) of this:

let $\alpha$ be the smallest angle opposite $2$ and $\gamma$ be the largest angle opposite $4$

Using the Cosine Rule you can show that $\cos\alpha=\frac 78$ and $\cos\gamma=-\frac 14$

Then using the identity $\sin^2\frac {\alpha}{2}=\frac 12(1-\cos\alpha)$, you can show that the obtuse angle which comprises $\alpha$ and its externally bisected part i.e $\frac{\pi+\alpha}{2}$ is given by $$\cos(\frac{\pi+\alpha}{2})=-\sin\frac{\alpha}{2}=-\frac 14$$

So this angle is $\gamma$ and similarity follows.