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# In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that: (i) $$\triangle$$ ABC ~ $$\triangle$$ AMP (ii) CA/PA = BC/MP

Given, ABC and AMP are two right triangles, right angled at B and M respectively.

(i) In $$\triangle$$ ABC and $$\triangle$$ AMP, we have,
$$\angle$$ CAB = $$\angle$$ MAP (common angles)
$$\angle$$ ABC = $$\angle$$ AMP = 90° (each 90°)
$$\triangle$$ ABC ~ $$\triangle$$ AMP (AA similarity criterion)

(ii) As, $$\triangle$$ ABC ~ $$\triangle$$ AMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP