$\mathrm{Az} A B C$ is a right-angled triangle, and we construct squares on its sides. The vertices of these squares - which are not the vertices of the triangle - form a hexagon. The area of this hexagon needs to be calculated! Known: the hypotenuse $(c)$ and the sum of the two legs $(d)$!
A. The vertices of the inscribed hexagon, according to our diagram, are $D, E, K, H, G, F$. The hexagon can be composed of the following parts:

![](https://cdn.mathpix.com/cropped/2024_05_02_6640e1db56942df59f48g-1.jpg?height=550&width=553&top_left_y=224&top_left_x=775)

1. $\triangle ABC$; its area is: $\frac{1}{2} AB \cdot AC = \frac{1}{2} cb$. (T. i. $AB=c, AC=b, BC=a$).
2. Quadrilateral $BCDE$; its area is $=a^2$,
3. Quadrilateral $ACFG$; its area is $=b^2$,
4. Quadrilateral $ABKH$; its area is $=c^2$.
5. $\triangle AHG$; this is congruent to $\triangle ABC$; its area is: $\frac{1}{2} bc$.
6. Triangles $BEK$ and $CDF$. Each of these triangles has the same area as $\triangle ABC$. Draw perpendiculars from $K$ and $A$ to $BE$ and its extension, $KN$ and $AM$. Then $\triangle ABM \cong \triangle BKN$, because their hypotenuses are equal: $AB = BK$ and, for example, $\angle KBN = \angle BAN$, because their sides are perpendicular to each other. This implies that $KN = BM$.

However, $KN$ is the height corresponding to side $BE$ in $\triangle BEK$, and $BM$ is the height corresponding to side $BC$ in $\triangle ABC$: $BE = BC$ and $KN = BM$, so the areas of the two triangles are equal.

Similarly, it can be shown that the area of $\triangle CDF$ is also equal to the area of $\triangle ABC$. Therefore, the area of the hexagon is:

$$
\frac{bc}{2} + a^2 + b^2 + c^2 + \frac{bc}{2} + 2 \frac{bc}{2} = a^2 + b^2 + c^2 + 2bc = a^2 + (b+c)^2 = a^2 + d^2
$$

Sándor Baka (Áll. Kemény Zsigmond g. V. o. Bp.)

Note. The area of $\triangle BEK$ is equal to the area of $\triangle ABC$; we could also have referred to the fact that these two triangles have two pairs of equal sides, i.e., $BK = BA$ and $BE = BC$, and the angles between these sides are supplementary. (For example, we get two such triangles when we draw the median in a triangle!)