To solve for $BD$ and $BC$, we will use Pythagoras' theorem and the geometric properties of the triangle.

Given:

• $AB = 12 \, \text{cm}$,
• $AD = 6 \, \text{cm}$,
• $CD = 18 \, \text{cm}$.

Step 1: Find $BD$ (using triangle $\triangle ABD$)

In $\triangle ABD$, $BD$ is perpendicular to $AC$. Using Pythagoras' theorem: $AB^2 = AD^2 + BD^2$ Substitute the known values: $12^2 = 6^2 + BD^2$ $144 = 36 + BD^2$ $BD^2 = 144 - 36 = 108$ $BD = \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3} \, \text{cm}.$

Step 2: Find $BC$ (using triangle $\triangle BCD$)

Now, $BC$ is the hypotenuse of $\triangle BCD$. Using Pythagoras' theorem: $BC^2 = BD^2 + CD^2$ Substitute the known values: $BC^2 = (6\sqrt{3})^2 + 18^2$ $BC^2 = 108 + 324 = 432$ $BC = \sqrt{432} = \sqrt{144 \cdot 3} = 12\sqrt{3} \, \text{cm}.$

Final Results:

• $BD = 6\sqrt{3} \, \text{cm}$,
• $BC = 12\sqrt{3} \, \text{cm}$.

Let me know if you need a detailed breakdown of the steps or have further questions. Here are some follow-up questions to deepen understanding:

1. How can you verify the perpendicularity of $BD$ using coordinates or slopes?
2. What are the properties of a right triangle that simplify these calculations?
3. Could trigonometric ratios be used to find these values instead?
4. How would the results change if the dimensions of $AB$, $AD$, and $CD$ were scaled proportionally?
5. Can you find the area of $\triangle ABC$ using the derived lengths?

Tip: Remember, using radicals (surds) helps keep results exact instead of