In △ABC, ∠B = 90° and BD ⊥ AC.

(i) If CD = 10 cm and BD = 8 cm; find AD.

(ii) If AC = 18 cm and AD = 6 cm; find BD.

(iii) If AC = 9 cm and AB = 7 cm; find AD.

△ABC is shown in the figure below:

(i) In △CDB,

⇒ ∠1 + ∠2 + ∠3 = 180° (Sum of angles of triangle = 180°)

⇒ ∠1 + ∠3 + 90° = 180°

⇒ ∠1 + ∠3 = 90° ……….(1)

From figure,

⇒ ∠B = 90°

⇒ ∠3 + ∠4 = 90° ……….(2)

From (1) and (2) we get,

⇒ ∠1 + ∠3 = ∠3 + ∠4

⇒ ∠1 = ∠4.

From figure,

⇒ ∠2 = ∠5 [Both = 90°]

∴ △CDB ~ △BDA [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\dfrac{CD}{BD} = \dfrac{BD}{AD}$ ……….(3)

Substituting values we get :

$\Rightarrow \dfrac{10}{8} = \dfrac{8}{AD} \\[1em] \Rightarrow AD = \dfrac{8^2}{10} \\[1em] \Rightarrow AD = \dfrac{64}{10} = 6.4 \text{ cm}.$

Hence, AD = 6.4 cm.

(ii) From figure,

CD = AC - AD = 18 - 6 = 12 cm.

Substituting values in (3) we get :

$\Rightarrow \dfrac{12}{BD} = \dfrac{BD}{6} \\[1em] \Rightarrow BD^2 = 12 \times 6 \\[1em] \Rightarrow BD^2 = 72 \\[1em] \Rightarrow BD = \sqrt{72} = 6\sqrt{2} = 8.5\text{ cm}.$

Hence, BD = 8.5 cm.

(iii) In △ABC and △ABD,

⇒ ∠ADB = ∠ABC [Both = 90°]

⇒ ∠ABD = ∠ACB [As ∠1 = ∠4]

∴ △ABC ~ △ABD [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{AD}{AB} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{AD}{7} = \dfrac{7}{9} \\[1em] \Rightarrow AD = 7 \times \dfrac{7}{9} = \dfrac{49}{9} = 5\dfrac{4}{9} \text{ cm}.$

Hence, AD = $5\dfrac{4}{9}$ cm.