To solve this problem, we need to analyze the right-angled triangle $\triangle ABC$ with the given conditions:

• $AC = 20$ cm (AC is a side of the right-angled triangle, let's assume it's the base).
• $CD = 10$ cm (CD is perpendicular to AB, forming another right-angled triangle $\triangle BCD$).

Steps to find the length of $BD$:

1. Interpret the problem:

   • Since $CD$ is perpendicular to $AB$, $D$ is the point on $AB$ such that $CD$ is a height from $C$ to $AB$. This creates two right-angled triangles: $\triangle ACD$ and $\triangle BCD$.

2. Use properties of similar triangles:

   $\triangle ACD$ and $\triangle BCD$ share the angle $\angle ACD$, and both are right triangles, meaning they are similar by AA (Angle-Angle) similarity.

3. Set up the ratio using similar triangles:

   • Since $\triangle ACD \sim \triangle BCD$, the sides of the triangles are proportional.

   • Let $AD = x$ and $BD = y$. So, $AB = AD + BD = x + y$.

4. Using the similarity ratios:

   From the similarity, we have: $\frac{AD}{AC} = \frac{CD}{BD}$

   Substituting the known values: $\frac{x}{20} = \frac{10}{y}$

   Cross-multiplying gives: $xy = 200$

   Additionally, since $D$ divides $AB$ into $AD$ and $BD$, and using the triangle similarity ratio $\frac{AD}{AC} = \frac{CD}{BD}$, we have $x = \frac{10}{y} \times 20$.

5. Solving for $BD$:

   We use the equation $xy = 200$ and let $AD = x$, $BD = y$.

   Since $AB = x + y$ and knowing the triangles' proportions, from $\triangle ABC$, we can use:

   $AC^2 = AD \cdot AB$

   $20^2 = x \cdot (x + y)$

   We know $xy = 200$, and $y = \frac{200}{x}$. Substituting this into $AB = x + y$:

   $400 = x \cdot (x + \frac{200}{x})$

   Simplify:

   $400 = x^2 + 200$

   $x^2 = 200$

   $x = \sqrt{200}$

   Since $x = 10\sqrt{2}$, we can find $BD = y$:

   $y = \frac{200}{x} = \frac{200}{10\sqrt{2}} = 10\sqrt{2}$

   Therefore, $BD = 10 \sqrt{2} \approx 14.14$ cm.

Conclusion: The length of $BD$ is $10 \sqrt{2}$ cm.

Would you like further details, or do you have any other questions?

Here are some related questions that you might find interesting:

1. How can we calculate the area of $\triangle ABC$?
2. What are the lengths of $AD$ and $CD$ using trigonometry?
3. How would the length of $BD$ change if $CD$ was 15 cm instead of 10 cm?
4. How can we find the hypotenuse $AB$ using the lengths we have?
5. What is the length of $AB$ directly using the Pythagorean theorem?

Tip: When dealing with similar triangles, remember that corresponding side lengths are proportional. This property is helpful in finding unknown sides.