1. **Stating the problem:** We have a right triangle ABC with AB = 15 cm (base) and AC = 10 cm (height). Inside it, a square AEFD is inscribed with vertex A common to the triangle, and points E, F, D lying on sides AB, BC, and AC respectively. 2. **Goal:** Find the side length of the square AEFD. 3. **Setup:** Let the side length of the square be $x$. 4. **Coordinates:** Place point A at origin $(0,0)$, AB along the x-axis, and AC along the y-axis. - Then $A = (0,0)$ - $B = (15,0)$ - $C = (0,10)$ 5. **Square vertices:** - $E$ lies on AB, so $E = (x,0)$ - $D$ lies on AC, so $D = (0,x)$ - $F$ lies on BC, so $F = (f_x,f_y)$ on line BC 6. **Equation of BC:** Line through $B(15,0)$ and $C(0,10)$ has slope $m = \frac{10-0}{0-15} = -\frac{2}{3}$. Equation: $$y - 0 = -\frac{2}{3}(x - 15) \implies y = -\frac{2}{3}x + 10$$ 7. **Point F on BC:** Since $F$ is on BC and also on the square, $F$ must be at $(x, y)$ where $y = -\frac{2}{3}x + 10$. 8. **Square properties:** The square AEFD has side length $x$. - $AE$ is horizontal of length $x$. - $AD$ is vertical of length $x$. - $EF$ and $DF$ are sides of the square, so $EF$ is vertical and $DF$ is horizontal. 9. **Coordinates of F:** Since $E = (x,0)$ and $F$ is vertically above $E$ by length $x$, then $F = (x, x)$. 10. **Check if $F$ lies on BC:** Substitute $x$ for $x$ and $x$ for $y$ in BC equation: $$x = -\frac{2}{3}x + 10$$ 11. **Solve for $x$:** $$x + \frac{2}{3}x = 10$$ $$\frac{3}{3}x + \frac{2}{3}x = 10$$ $$\frac{5}{3}x = 10$$ $$x = 10 \times \frac{3}{5} = 6$$ 12. **Conclusion:** The side length of the square AEFD is $6$ cm.