In the figure,$\Delta ABC$ and $\Delta AMP$ are two right-angled triangles,right-angled at $B$ and $M$ respectively. Prove that:
$(i)$ $\Delta ABC \sim \Delta AMP$
$(ii)$ $\frac{CA}{PA} = \frac{BC}{MP}$
$(i)$ $\Delta ABC \sim \Delta AMP$
$(ii)$ $\frac{CA}{PA} = \frac{BC}{MP}$
(N/A) In $\Delta ABC$ and $\Delta AMP$:
$\angle ABC = \angle AMP = 90^{\circ}$ (Given)
$\angle A = \angle A$ (Common angle)
Therefore,$\Delta ABC \sim \Delta AMP$ by $AA$ similarity criterion.
Since the triangles are similar,their corresponding sides are proportional.
Thus,$\frac{CA}{PA} = \frac{BC}{MP}$.
$\angle ABC = \angle AMP = 90^{\circ}$ (Given)
$\angle A = \angle A$ (Common angle)
Therefore,$\Delta ABC \sim \Delta AMP$ by $AA$ similarity criterion.
Since the triangles are similar,their corresponding sides are proportional.
Thus,$\frac{CA}{PA} = \frac{BC}{MP}$.
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