In the given figure, \(ABC\) and \(AMP\) are two right triangles, right angled at \(B\) and \(M\) respectively. Prove that:
(i) \(\Delta ABC \sim \Delta AMP\)
(ii) \(\frac{CA}{PA} = \frac{BC}{MP}\)
Proof:
(i) Given \(ABC\) and \(AMP\) are two right triangles, \(\angle ABC = 90^\circ\), \(\angle AMP = 90^\circ\)
In \(\Delta ABC\) and \(\Delta AMP\),
\(\angle CAB = \angle\) (Common angle)
\(\angle ABC = \angle\) (Both \(90^\circ\))
Thus, \(\Delta ABC \sim \Delta AMP\) (\(AA\) similarity)
(ii) In the first part we proved, \(\Delta ABC\) and \(\Delta AMP\)
If two triangles are similar, then the ratio of their corresponding sides is proportional.
\(\frac{CA}{PA} = \frac{BC}{MP}\)
Hence proved.
Answer variants:
MAP
AMP
\(\frac{CA}{PA} = \frac{BC}{MP} = \frac{AB}{AM}\)