Question.
In figure, $\mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively. Prove that:
(i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$
(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
In figure, $\mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively. Prove that:
(i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$
(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
Solution:
(i) In $\Delta \mathrm{ABC}$ and $\Delta \mathrm{AMP}$
$\angle \mathrm{CAB}=\angle \mathrm{PAM}($ common $)$
$\angle \mathrm{ABC}=\angle \mathrm{AMP}=90^{\circ}$
$\therefore$ By AA similarity
$\Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$
(ii) As $\triangle \mathrm{ABC} \sim \Delta \mathrm{AMP}$ (Proved above)
$\therefore \quad \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
(i) In $\Delta \mathrm{ABC}$ and $\Delta \mathrm{AMP}$
$\angle \mathrm{CAB}=\angle \mathrm{PAM}($ common $)$
$\angle \mathrm{ABC}=\angle \mathrm{AMP}=90^{\circ}$
$\therefore$ By AA similarity
$\Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$
(ii) As $\triangle \mathrm{ABC} \sim \Delta \mathrm{AMP}$ (Proved above)
$\therefore \quad \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
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