Question - Exploring Similarity and Area Ratios in Geometric Figures

(a) Triangle \( AXM \) is similar to triangle \( CXD \) because they have two pairs of corresponding angles that are equal: \( \angle AXM = \angle CXD \) (both are right angles), and \( \angle AMX = \angle CDX \) (they are vertical angles). Therefore, by AA (Angle-Angle) similarity criterion, \( \triangle AXM \sim \triangle CXD \).

(b) To find the value of \( \frac{\text{area of } \triangle AMX}{\text{area of } \triangle CXD} \), we can use the fact that the ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. If \( AM = x \) and \( CD = kx \) for some proportional constant \( k \), then:

\[ \frac{\text{area of } \triangle AMX}{\text{area of } \triangle CXD} = \left( \frac{x}{kx} \right)^2 = \left( \frac{1}{k} \right)^2 \]

(c) By the same principle, to find the value of \( \frac{\text{area of } \triangle AXM}{\text{area of } \triangle ACDX} \), we add the areas of \( \triangle AMX \) and \( \triangle CXD \) to get the area of \( \triangle ACDX \). Assuming \( AM = x \) and \( CD = kx \), the area of \( \triangle ACDX \) is the sum of the areas of \( \triangle AMX \) and \( \triangle CXD \):

\[ \text{area of }\triangle ACDX = \text{area of }\triangle AMX + \text{area of }\triangle CXD \]

\[ \frac{\text{area of }\triangle AXM}{\text{area of }\triangle ACDX} = \frac{\text{area of }\triangle AXM}{\text{area of }\triangle AXM + \text{area of }\triangle CXD} \]

\[ \frac{\text{area of }\triangle AXM}{\text{area of }\triangle ACDX} = \frac{x^2}{x^2 + (kx)^2} \]

Without specific values for AM and CD, we can't simplify further. If more information is provided, then the ratio can be expressed in simplest form.