Question - Geometric Similarity and Ratio Calculation in Triangles

(a) In triangle ABM and CDM:

\[\angle MAB = \angle MCD\] (Vertically opposite angles are equal)

\[\angle AMB = \angle CMD\] (Each equals \(90^\circ\))

Therefore, by AA (Angle-Angle) similarity criterion, \(\triangle ABM \sim \triangle CDM\).

(b) The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides:

\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle CDM}} = \left(\frac{AM}{CM}\right)^2\]

(c) \[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \frac{\frac{1}{2} AM \times BM}{\frac{1}{2} AC \times BC}\]

Given that \(BM = MC\) and \(AM = MC\),

\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \left(\frac{MC}{AC}\right) \times \left(\frac{AM}{BC}\right)\]

Since \(AB = AC\), \(BC = AB - AM\),

\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \frac{MC}{AC} \times \frac{AM}{AB - AM}\]