A square in a triangle

We note the right $\triangle ABC$ is the popular right triangle with integer side lengths of ratio $3:4:5$ or $0.6:0.8:1$ . We also note that $\triangle AIJ$ and $\triangle BIH$ are similar to $\triangle ABC$ .

Let the side length of the square be $a$ , then $AI = 0.6 a$ and $IB = \dfrac a{0.8}$ . Since:

$\begin{aligned} AI + IB & = AB \\ 0.6a + \frac a{0.8} & = 6 \\ \frac {0.48a+a}{0.8} & = 6 \\ \implies a & = \frac {4.80}{1.48} = \boxed{\frac {120}{37}} \end{aligned}$

Let $\angle {ACB}=α$ . Then $\tan α=\dfrac{3}{4}$ .

Let the length of each side of the square be $a$ . Then

$\left (\dfrac{4}{3}+1+\dfrac{3}{4}\right ) a=\sqrt {6^2+8^2}=10$

$\implies a=\boxed {\dfrac{120}{37}}$ .

I have a derived formula about problems like this. Let $x$ be the side length of the square. Then,

$x = \dfrac{bh}{b+h}$ where: $b$ is the side of the triangle where the side of the square coincides and $h$ is the perpendicular distance from $b$ to the opposite vertex.

Note that $\triangle ABC$ is the popular pythagorean triple, $3-4-5$ right triangle. So the length of $BC$ is $10$ . Now we need to find $h$ ,

$\sin \angle ABC = \dfrac{AC}{BC}=\dfrac{h}{AB}$

$\dfrac{8}{10}=\dfrac{h}{6}$

$h = \dfrac{24}{5}$

Now, substitute.

$x = \dfrac{bh}{b+h}=\dfrac{10 \times \dfrac{24}{5}}{10 + \dfrac{24}{5}}=\color{#3D99F6}\boxed{\dfrac{120}{37}}$ $\color{#69047E}\boxed{answer}$

$\begin{aligned} & BC = 10 {\color{#D61F06} \text{ (pythagorean theorem)}} \\ & let \ HG = x {\color{#D61F06} =HI=GJ}\\ & \triangle ABC \sim \triangle HBI \Rightarrow \frac{AB}{HB} = \frac{AC}{HI}\Rightarrow HB=\frac{3}{4} x \\ & \triangle ABC \sim \triangle GCJ \Rightarrow \frac{AB}{GC} = \frac{AC}{GJ}\Rightarrow GC=\frac{4}{3} x \\ & BC = BH + HG + GC \Rightarrow 10 = \frac{37}{12} x \Rightarrow x=\frac{120}{37} \end{aligned}$