This note gives the common calculations in the solution for problems titled Dynamic Geometry: P96, 101, 106, 109, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, and 125 so far by @Valentin Duringer. .

Let the chord dividing a circle with center at $O$ into two circular segments of heights $4$ and $9$ be $AB$. Diameter $DE$ through $O$ cut $AB$ perpendicularly at $F$. Then $DF = 4$, $FE = 9$, and diameter $DE = DF + FE = 13$, therefore radius of the circle is $6.5$. By intersecting chords theorem $AF \cdot FB = DF \cdot FE = 4 \cdot 9 \implies AF = FB = 6 \implies AB=12$. If we set $O$ as the origin $(0,0)$ of the $xy$-plane with $AB$ parallel to the $x$-axis, then $AB$ is along $y=2.5$, and $A = (-6, 2.5)$ and $B=(6,2.5)$.

Label the upper variable triangle as $ABC$ and the lower variable triangle $ABC'$. Since the inscribed angle $\angle AC'B$ and the central angle $\angle AOB$ both form the same arc $ACB$, $\angle AC'B = \frac 12 \angle AOB = \frac 12 \left(2 \tan^{-1} \frac {12}5 \right) = \tan^{-1} \frac {12}5$. Similarly, $\angle ACB = \pi - \tan^{-1} \frac {12}5$.

By sine rule:

$\frac {AC}{AB} = \frac {\sin \angle CBA}{\sin \angle ACB} \implies AC = \frac {12\angle CBA}{\frac {12}{13}} = 13\sin \angle CBA$

Similarly, $BC = 13 \sin \angle CAB$, $AC' = 13\sin \angle C'BA$, and $BC' = 13 \sin \angle C'AB$,

Consider any $\triangle ABC$ with a height $CD = h$ and a width of $AB=w$. Let the rectangle $\triangle ABC$ inscribes be $KLMN$ and the height of $\triangle CKL$ be $CE = \alpha h$, where $0 \le \alpha \le 1$. Then the area of rectangle $KLMN$, $A = KL \cdot LM$. Since $\triangle CKL$ and $\triangle ABC$ are similar $KL = \alpha \cdot AB = \alpha w$. And $LM = CD - CE = h - \alpha h$. Then $A = \alpha (1-\alpha) wh$. By AM-GM inequality, $2\sqrt{\alpha(1-\alpha)} \le \alpha + (1 - \alpha) = 1$, and equality occurs when $\alpha = 1 - \alpha \implies \alpha = \frac 12$. Then the rectangle has a maximum area $A_{\max} = \frac 14 wh$, when the height and breadth of the rectangle are half of that of the triangle.

Let us consider the relationship between the side length $s$ of a square $KLMN$ to the height $CD=h$ and width $AB=w$ of the triangle $\triangle ABC$ that inscribes it. Note that $\triangle KLC$ and $\triangle ABC$ are similar. Let the height of $\triangle KLC$ be $CE$, then $CE = \dfrac swh$. From $CD = CE + ED$, we have

$h = \frac swh + s \implies s = \frac h{1+\frac hw} = \frac {wh}{h+w} = \frac {12h}{h+12} \quad \small \blue{\text{Since }w=AB=12}$

Similarly for the square in $\triangle ABC'$, $s' = \dfrac {12h'}{h'+12}$.

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TopNewestNice work sir !

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Actually, the coordinate system I use is different, I use point A(0,0)

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