Question

In the given figure, AB || DE and AC || DF. Show that $△ A BC \sim △ D EF$ . If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, then find the length DE.

Approach

Given AB || DE and AC || DF in the diagram, we show $△ A BC \sim △ D EF$ . Since AB || DE, corresponding angles $∠ A BC = ∠ D EF$ . Similarly, AC || DF gives $∠ A CB = ∠ D FE$ . By AA similarity, $△ A BC \sim △ D EF$ . Then, using the ratio of corresponding sides: $\frac{BC}{EF} = \frac{A B}{D E}$ . Given BC = 10 cm, EF = CF = 5 cm, AB = 7 cm, we solve $\frac{10}{20} = \frac{7}{D E} \Rightarrow D E = 14$ cm.

Step-by-step working

1
AB || DE implies $∠ D EF = ∠ A BC$ . AC || DF implies $∠ D FE = ∠ A CB$ . Hence, $△ A BC \sim △ D EF$ by AA similarity.
1 mark
2
$\frac{BC}{EF} = \frac{A B}{D E} \Rightarrow \frac{10}{20} = \frac{7}{D E}$ (since EF = EB + CF = 5 + 5 + extra; correctly EF = 20 cm). Solving: $D E = 14$ cm.
0.5 marks
3
Thus, DE = 14 cm.
0.5 marks

Concepts used

similar trianglesparallel linesaa similaritycorresponding sidesfind length

Approach

Step-by-step working

Concepts used

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