Triangle Similarity Area Ratio

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. This fundamental property is used to solve the problem.

Similarity and Corresponding Sides

The problem states that $\triangle ABC \sim \triangle DBA$. This similarity implies a specific correspondence between the vertices and sides of the two triangles:

• Vertex A corresponds to Vertex D.
• Vertex B corresponds to Vertex B.
• Vertex C corresponds to Vertex A.

Based on this vertex correspondence, the ratios of the corresponding sides are:

$ \frac{AB}{DB} = \frac{BC}{BA} = \frac{AC}{DA} $

Calculating the Area Ratio

We are given the ratio $AB : BC = m : n$. This can be written as a fraction:

$ \frac{AB}{BC} = \frac{m}{n} $

The ratio of the areas is given by the square of the ratio of any pair of corresponding sides. We use the sides BC (from $\triangle ABC$) and BA (from $\triangle DBA$):

$ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DBA)} = \left( \frac{BC}{BA} \right)^2 $

Since $BA$ is the same as $AB$, the equation becomes:

$ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DBA)} = \left( \frac{BC}{AB} \right)^2 $

From the given ratio $\frac{AB}{BC} = \frac{m}{n}$, we can find the inverse ratio $\frac{BC}{AB}$:

$ \frac{BC}{AB} = \frac{n}{m} $

Now, substitute this value into the area ratio formula:

$ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DBA)} = \left( \frac{n}{m} \right)^2 = \frac{n^2}{m^2} $

Since the area of $\triangle DBA$ is the same as the area of $\triangle ABD$, the ratio $\triangle$ ABC : $\triangle$ ABD is $n^2 : m^2$.