The given diagram features two right-angled triangles sharing side $AC$, and we need to solve for the length of $AD$. Let's break it down:

Known information:

1. Triangle $ABC$ is a right-angled triangle at $C$.

   • Angle $\angle ABC = 30^\circ$
   • Side $BC = 10$ cm
   • Angle $\angle BAC = 45^\circ$

2. Triangle $ACD$ also shares side $AC$ and forms a right angle at $C$.

   • Side $CD = 5$ cm

Steps to solve for $AD$:

1. Find $AC$ using triangle $ABC$:

Since $ABC$ is a right triangle, we can use trigonometric relationships to find $AC$. Specifically, use the sine function:

$\sin(30^\circ) = \frac{AC}{BC}$

$AC = BC \cdot \sin(30^\circ) = 10 \cdot 0.5 = 5 \, \text{cm}$

2. Find $AD$ using triangle $ACD$:

Now that we know $AC = 5$ cm and $CD = 5$ cm, triangle $ACD$ is an isosceles right triangle (since $\angle ACD = 90^\circ$). We can use the Pythagorean theorem to find $AD$:

$AD^2 = AC^2 + CD^2$

$AD^2 = 5^2 + 5^2 = 25 + 25 = 50$

$AD = \sqrt{50} = 5\sqrt{2} \, \text{cm} \approx 7.07 \, \text{cm}$

Thus, the length of $AD$ is approximately 7.07 cm.

Let me know if you'd like further clarification or details! Here are some related questions to explore:

1. What other methods can you use to solve for $AD$?
2. How would the solution change if $CD$ was a different length?
3. Can you derive the exact value of $\sin(45^\circ)$ and $\sin(30^\circ)$?
4. How can you verify the result using the Pythagorean theorem?
5. What happens to the angles if triangle $ACD$ were not isosceles?

Tip: Always recheck your intermediate calculations to ensure that the final result is consistent.