In the given figure, ΔABC and ΔAMP are right angled at B and M respectively.

Given AC = 10 cm, AP = 15 cm and PM = 12 cm.

(i) Prove that : ∆ABC ~ ∆AMP.

(ii) Find AB and BC.

(i) In ∆ABC and ∆AMP, we have

⇒ ∠BAC = ∠PAM [Common]

⇒ ∠ABC = ∠PMA [Each = 90°]

∴ ∆ABC ~ ∆AMP [By AA]

Hence, proved that, ∆ABC ~ ∆AMP.

(ii) In right angle triangle AMP,

By pythagoras theorem,

⇒ AP² = AM² + MP²

⇒ AM² = AP² - MP²

⇒ AM² = 15² - 12²

⇒ AM² = 225 - 144

⇒ AM² = 81

⇒ AM = $\sqrt{81}$ = 9 cm.

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

$\Rightarrow \dfrac{AB}{AM} = \dfrac{AC}{AP} \\[1em] \Rightarrow \dfrac{AB}{9} = \dfrac{10}{15} \\[1em] \Rightarrow AB = 9 \times \dfrac{10}{15} \\[1em] \Rightarrow AB = 6 \text{ cm}.$

Also,

$\Rightarrow \dfrac{BC}{MP} = \dfrac{AC}{AP} \\[1em] \Rightarrow \dfrac{BC}{12} = \dfrac{10}{15} \\[1em] \Rightarrow BC = 12 \times \dfrac{10}{15} \\[1em] \Rightarrow BC = 8 \text{ cm}.$

Hence, AB = 6 cm and BC = 8 cm.