QQuestion
In the given diagram, DE ∥ BC and AD : DB = 2 : 3.
(a) Prove that: ΔADE ~ ΔABC and hence find DE : BC.
(b) Prove: ΔDFE ~ ΔCFB.
(c) Given, area of ΔDFE = 16 square units, find the area of ΔCFB.
ICSEClass XMathematicsSimilarity04 Mark2026 Easy
✓Answer
(a) In ΔADE and ΔABC:
∠ADE = ∠ABC (corresponding angles)
∠AED = ∠ACB (corresponding angles)
Therefore, ΔADE ~ ΔABC by AA similarity.
Given AD : DB = 2 : 3.
Let AD = 2x and DB = 3x.
Then AB = AD + DB = 5x.
\frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{2\mathrm{x}}{5\mathrm{x}}=\frac{2}{5}Therefore, DE : BC = 2 : 5.
(b) In ΔDFE and ΔCFB:
∠DFE = ∠CFB (vertically opposite angles)
∠DEF = ∠CBF (alternate interior angles)
Therefore, ΔDFE ~ ΔCFB by AA similarity.
(c) Let area of ΔCFB = x square units.
⇒ \frac{16}{\mathrm{x}}=\left(\frac{2}{5}\right)^2=\frac{4}{25}
⇒ 4x = 16 × 25
⇒ x = \frac{16\times25}{4} = 100
Therefore, area of ΔCFB = 100 square units.
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