Problem
What is the length of one side of a square inscribed in a triangle, of which the longest side is $21 \ inches$ and the perpendicular line from the vertex to the longest side is $8 \ inches$?
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Solution
Let the lengths of the longest side $BC$ of a triangle $ABC$, the perpendicular line $AH$ drawn from the vertex to the longest side and one side of the square be $a, \ b$ and $x$, respectively.
Then, for similar right triangles $ADI$ and $DBE$,
$$AI∶DE=DI∶BE,$$
$$(b-x)∶x=DI∶BE,$$
$$∴ \ DI=\frac{b-x}{x}×BE.$$
Similarly, for similar right triangles $AIG$ and $GFC$,
$$AI∶GF=IG∶FC,$$
$$(b-x)∶x=IG∶FC,$$
$$∴ \ IG=\frac{b-x}{x}×FC.$$
Since $DG=DI+IG=x$,
$$\frac{b-x}{x}×(BE+FC)=x,$$
$$(b-x)(BE+FC)=x^2,$$
$$∴ \ x^2+(BE+FC)x-(BE+FC)b=0. \qquad [1]$$
However, since $BC=BE+EF+FC=a$,
$$BE+FC=a-x. \qquad [2]$$
From $[1]$ and $[2]$,
$$x^2+(a-x)x-(a-x)b=0,$$
$$(a+b)x-ab=0,$$
$$∴ \ x=\frac{ab}{a+b}. \qquad [3]$$
Substituting $a=21$ and $b=8$ into $[3]$,
$$x=\frac{21×8}{21+8}=\frac{168}{29}≒5.793103448275862.$$
$$(Answer) \quad approximately \quad 5.79310 \quad inches.$$
Note
In the original text, the answer is $^”5.70931^”$.
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.27; pp.399-400.