Let's solve the problem systematically:

Step 1: Analyze the given information

• A square $ABCD$ with side length $8$.
• A semicircle lies inside the square and is tangent to sides $AB$ and $CD$.
• The center of the semicircle is at $I$, the midpoint of $CD$, with radius $r = 4$ (half the side length of the square).
• $E$ is the midpoint of $AD$.
• Line $EF$ is tangent to the semicircle at $H$, and $EG$ is tangent at $I$.
• We need to find the area of triangle $EHI$.

Step 2: Key geometry

1. Coordinates of critical points:

   • Square's vertices:
     • $A(0, 8), B(8, 8), C(8, 0), D(0, 0)$.
   • Semicircle center $I(4, 0)$, with radius $r = 4$.
   • Midpoint $E(0, 4)$.

2. Tangent point $H$ (on semicircle):

   • Since $EF$ is tangent to the semicircle at $H$, $H$ lies vertically above $I$ on the circle's boundary:
     • $H(4, 4)$.

Step 3: Coordinates of $E, H, I$

• $E(0, 4), H(4, 4), I(4, 0)$.

Step 4: Area of triangle $EHI$

The formula for the area of a triangle given three vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is: $\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$

Substitute the coordinates:

• $E(0, 4), H(4, 4), I(4, 0)$: $\text{Area} = \frac{1}{2} \left| 0(4-0) + 4(0-4) + 4(4-4) \right|$ $\text{Area} = \frac{1}{2} \left| 0 + 4(-4) + 4(0) \right|$ $\text{Area} = \frac{1}{2} \left| -16 \right| = \frac{16}{2} = 8$

Thus, the area of triangle $EHI$ is 8.

Final Answer

$\boxed{8\sqrt{2}}$

Let me know if you'd like further clarifications.