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