Solved 2023 Question Paper ICSE Class 10 Mathematics — Question 6

(a) Given,

DE // BF

∴ ∠DEG = ∠GBF (Alternate angles are equal)

∠EDG = ∠GFB (Alternate angles are equal)

⇒ △GED ~ △ GBF (By A.A. axiom)

Hence, proved that △GED ~ △ GBF.

(b) We know that,

Corresponding sides of similar triangle are proportional.

Since, △ GED ~ △ GBF

$\Rightarrow \dfrac{DE}{BF} = \dfrac{EG}{BG} \\[1em] \Rightarrow \dfrac{DE}{30} = \dfrac{8}{16} \\[1em] \Rightarrow DE = 30 \times \dfrac{8}{16} \\[1em] \Rightarrow DE = 15 \text{ cm}.$

Hence, DE = 15 cm.

(c) In △ BAC and △ BDE,

⇒ ∠BAC = ∠BDE (Corresponding angles are equal)

⇒ ∠ABC = ∠DBE (Common angle)

⇒ △ BAC ~ △ BDE (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional.

Since, △ BAC ~ △ BDE

$\Rightarrow \dfrac{AB}{BD} = \dfrac{AC}{DE} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{DE}{AC} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{15}{24} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{5}{8}$

⇒ BD : AB = 5 : 8.

Hence, BD : AB = 5 : 8.