Given: Δ ABC such that AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm. Also DE ∥ BC.
Required to find: BD and CE.
As DE ∥ BC, AB is transversal,
∠APQ = ∠ABC (corresponding angles)
As DE ∥ BC, AC is transversal,
∠AED = ∠ACB (corresponding angles)
In Δ ADE and Δ ABC,
∠ADE = ∠ABC
∠AED = ∠ACB
∴ Δ ADE = Δ ABC (AA similarity criteria)
Now, we know that
Corresponding parts of similar triangles are propositional.
⇒ \(\frac{AD}{AB}\) = \(\frac{AE}{AC}\) = \(\frac{DE}{BC}\)
\(\frac{AD}{AB}\) = \(\frac{DE}{BC}\)
\(\frac{2.4}{(2.4 + DB)}\) = \(\frac{2}{5}\) [Since, AB = AD + DB]
2.4 + DB = 6
DB = 6 – 2.4
DB = 3.6 cm
In the same way,
⇒ \(\frac{AE}{AC} = \frac{DE}{BC}\)
\(\frac{3.2}{(3.2 + EC)}\) = \(\frac{2}{5}\) [Since AC = AE + EC]
3.2 + EC = 8
EC = 8 – 3.2
EC = 4.8 cm
∴ BD = 3.6 cm and CE = 4.8 cm.
