Geometric Similarity and Ratio Calculation in Triangles
<p>(a) In triangle ABM and CDM:</p>
<p>\[\angle MAB = \angle MCD\] (Vertically opposite angles are equal)</p>
<p>\[\angle AMB = \angle CMD\] (Each equals \(90^\circ\))</p>
<p>Therefore, by AA (Angle-Angle) similarity criterion, \(\triangle ABM \sim \triangle CDM\).</p>
<p>(b) The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides:</p>
<p>\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle CDM}} = \left(\frac{AM}{CM}\right)^2\]</p>
<p>(c) \[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \frac{\frac{1}{2} AM \times BM}{\frac{1}{2} AC \times BC}\]</p>
<p>Given that \(BM = MC\) and \(AM = MC\),</p>
<p>\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \left(\frac{MC}{AC}\right) \times \left(\frac{AM}{BC}\right)\]</p>
<p>Since \(AB = AC\), \(BC = AB - AM\),</p>
<p>\[\frac{{\text{area of } \triangle ABM}}{{\text{area of } \triangle ABC}} = \frac{MC}{AC} \times \frac{AM}{AB - AM}\]</p>
