In figure, $\mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles,
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) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$

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)

$\therefore \quad \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
Editor