Question

In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

Hint:

Always double-check the segment additions along a straight line transversal.
Calculating $EF = 20\text{ cm}$ correctly is key, as using $BC$ or $EB$ incorrectly in the ratio would lead to erroneous scale factors!

Answer 1

Step 1: Prove similarity using alternate angles instead of corresponding angles.
Since $E,B,C,F$ are collinear and $AB \parallel DE$, treating $BC$ as a transversal gives $\angle ABC = \angle DEF$ (corresponding angles on the parallel lines). Similarly $AC \parallel DF$ gives $\angle ACB = \angle DFE$. By AA similarity, $\triangle ABC \sim \triangle DEF$.
Step 2: Find the full length $EF$.
\[ EF = EB + BC + CF = 5+10+5 = 20\text{ cm} \]
Step 3: Use the similarity ratio to find $DE$, cross-multiplying directly.
\[ \frac{AB}{DE} = \frac{BC}{EF} \implies DE = \frac{AB \times EF}{BC} = \frac{7 \times 20}{10} \]
Step 4: Simplify and conclude.
\[ DE = 14\text{ cm} \]
Triangles $ABC$ and $DEF$ are similar, and $DE = 14\text{ cm}$.
\[ \boxed{DE = 14\text{ cm}} \]