Question - Similarity and Ratio in a Parallelogram

\text{(a) Similarity Reasoning:}

Triangle \triangle AXM \:\text{is similar to}\: \triangle CXD \:\text{because}:

\text{1. } \angle AXM = \angle CXD \:\text{opposite angles are equal in a parallelogram}.

\text{2. } \angle A = \angle C \:\text{alternate angles are equal as AD} \parallel \text{BC in a parallelogram}.

\text{Therefore, by AA similarity criterion, } \triangle AXM \sim \triangle CXD

\text{(b) Area Ratio Calculation:}

\text{Since } \triangle AXM \sim \triangle CXD,\: \text{the ratio of their areas is the square of the ratio of their corresponding sides.}

\text{The corresponding sides are } AM \:\text{and} \: CD.\:

AM = \frac{1}{2}AD \:\text{since M is the midpoint of AD}.

CD = AD \:\text{as opposite sides of a parallelogram are equal}.

\text{Therefore, the ratio of } AM \:\text{to} \: CD \:\text{is} \:\frac{AM}{CD} = \frac{1/2 \cdot AD}{AD} = \frac{1}{2}.

\text{The ratio of areas is } (\frac{AM}{CD})^2 = (\frac{1}{2})^2 = \frac{1}{4}.

\text{Hence,} \:\frac{\text{area of } \triangle AXM}{\text{area of } \triangle CXD} = \frac{1}{4}.