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 the similarity using parallel lines as transversal pairs.
Since $E, B, C, F$ lie on one straight line, treat this line as a transversal. Because $AB \parallel DE$, the transversal $EBCF$ gives $\angle ABC = \angle DEF$ as corresponding angles. Because $AC \parallel DF$, the same transversal gives $\angle ACB = \angle DFE$.
Step 2: Apply the AA criterion.
With two pairs of equal angles, $\Delta ABC \sim \Delta DEF$ by AA similarity.
Step 3: Find the full length of EF.
Since the points lie in the order $E, B, C, F$ on the line, $EF = EB + BC + CF = 5 + 10 + 5 = 20\text{ cm}$.
Step 4: Use the similarity ratio to find DE.
Corresponding sides of similar triangles give $\frac{AB}{DE} = \frac{BC}{EF}$, so $\frac{7}{DE} = \frac{10}{20} = \frac{1}{2}$, which gives $DE = 14\text{ cm}$.
\[ \boxed{DE = 14\text{ cm}} \]