Square in Right-angled Triangle

Figure Suppose the side length of square $BHIJ$ is $x$ , of square $KLMN$ is $y$ .

In square $BHIJ$ , $\angle BJI=90\degree$ $\therefore \angle AJI=180\degree - \angle BJI =180\degree-90\degree=90\degree \, \because \angle ABC =90\degree \, \therefore \angle ABC =\angle AJI$

$\because \angle BAC =\angle BAC \, \therefore \triangle AJI \sim \triangle ABC \, \therefore \dfrac{AJ}{AB} =\dfrac{IJ}{BC} \, \therefore \dfrac{4-x}{4} =\dfrac{x}{3} \, \therefore x= \dfrac{12}{7}$

Made $OE \perp DF$ , and the junction of $OE$ and $DF$ is $O$ , of $OE$ and $KL$ is $P$ .

In right-angled triangle $Rt\triangle DEF$ , from Pythagorean theorem, easily gets $DF=\sqrt{DE^2+DF^2}=\sqrt{3^2+4^2}=5$

In right-angled triangle $Rt\triangle DEF$ , from the area, $S_{\triangle DEF}=\dfrac {DE \cdot DF}{2}=\dfrac{DF \cdot OE}{2}\, \therefore \dfrac{5\cdot OE}{2}=\dfrac{3\times4}{2}\,\therefore OE=\dfrac{12}{5}$ .

In square $KLMN$ , $OP=KL=y,\, KL \parallel MN \, \therefore KL \parallel DF \, \therefore \angle EDF =\angle EKL$ .

$\because \angle DEF=\angle DEF \, \therefore \triangle EKL \sim \triangle EDF \, \therefore \dfrac{EP}{OE}=\dfrac{KL}{DF} \, \therefore \dfrac{\dfrac{12}{5}-y}{\dfrac{12}{5}}=\dfrac{y}{5} \, \therefore y=\dfrac{60}{37}$

$\because x=\frac{12\times 5}{7\times5}=\frac{60}{35}>\frac{60}{37} \, \therefore x>y$ $\large{\therefore \text{Area}_{\text{Square BHIJ}} > \text{Area}_{\text{Square KLMN}}}_\square$

We note that all the seven triangles involved are $3$ - $4$ - $5$ right triangles.

Let the side length of square $BHIJ$ be $a$ . Since $\triangle AJI$ is a $3$ - $4$ - $5$ right triangle, $\dfrac {AJ}{JI} = \dfrac {4-a}a = \dfrac 43 \implies 12 -3a = 4a \implies a = \dfrac {12}7$ .

Similarly, let the side length of square $KLMN$ be $b$ , Then $KE=\dfrac 45b$ and $DK = 4-\dfrac 45 b$ and $\dfrac {DK}{KN} = \dfrac {4-\frac 45 b}b = \dfrac 53 \implies 12 - \dfrac {12}5b = 5b \implies b = \dfrac {60}{37} < \dfrac {60}{35} = \dfrac {12}7 = a$ .

Therefore, $\boxed{[BHIJ] < [KLMN]}$ .