2005 AMC 10A Problem 25

Compare $\triangle ADE$ with $\triangle ABC$ in two steps, each time changing one side along a line through $A$.

Draw $BE$. Triangles $ADE$ and $ABE$ have the same apex $E$ and bases $AD$, $AB$ on the same line, so
$$\frac{[ADE]}{[ABE]} = \frac{AD}{AB} = \frac{19}{25}.$$
Triangles $ABE$ and $ABC$ have the same apex $B$ and bases $AE$, $AC$ on the same line, so
$$\frac{[ABE]}{[ABC]} = \frac{AE}{AC} = \frac{14}{42} = \frac{1}{3}.$$
Multiplying,
$$\frac{[ADE]}{[ABC]} = \frac{19}{25} \cdot \frac{1}{3} = \frac{19}{75}.$$

Think of $[ABC]$ as $75$ units. Then $[ADE] = 19$ units and the quadrilateral $BCED$, which is everything else, has $75 - 19 = 56$ units. The requested ratio is
$$\frac{[ADE]}{[BCED]} = \frac{19}{56}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{19}{56}}$.

Two triangles that share an angle have areas in the ratio of the products of the sides enclosing that angle (the $\frac{1}{2}ab\sin\theta$ formula, or two applications of "same height, compare bases"). The side $BC = 39$ is never needed; it is there to tempt Heron's formula. Once a part-to-whole ratio is known, part-to-remainder is a subtraction.

Trig form: $[ADE] = \frac{1}{2}\cdot 19 \cdot 14 \sin A$ and $[ABC] = \frac{1}{2}\cdot 25 \cdot 42 \sin A$; the ratio $\frac{19 \cdot 14}{25 \cdot 42} = \frac{19}{75}$ appears immediately, and $\frac{19}{75 - 19} = \frac{19}{56}$.

• Reporting [ADE]/[ABC] = 19/75 (choice (B)) instead of the ratio to the quadrilateral, or trying to use BC = 39, which is irrelevant.
• Assuming $DE \parallel BC$ and squaring a single ratio; the ratios $\frac{19}{25}$ and $\frac{14}{42}$ differ, so the triangles are not similar.