In figure, altitudes $\mathrm{AD}$ and $\mathrm{CE}$ of $\triangle \mathrm{ABC}$ intersect each other at the point $\mathrm{P}$.

Solution:

(i) In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{CDP}$,

$\angle \mathrm{APE}=\angle \mathrm{CPD}$ (vertically opposite angles)

$\angle \mathrm{AEP}=\angle \mathrm{CDP}=90^{\circ}$

$\therefore \quad$ By AA similarity

$\Delta \mathrm{AEP} \sim \Delta \mathrm{CDP}$

(ii) In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{CBE}$,

$\angle \mathrm{ABD}=\angle \mathrm{CBE}($ common $)$

$\angle \mathrm{ADB}=\angle \mathrm{CEB}=90^{\circ}$

$\therefore$ By AA similarity

$\Delta \mathrm{ABD} \sim \Delta \mathrm{CBE}$

(iii) In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{ADB}$,

$\angle \mathrm{PAE}=\angle \mathrm{DAB}($ common $)$

$\angle \mathrm{AEP}=\angle \mathrm{ADB}=90^{\circ}$

$\therefore$ By AA similarity

$\Delta \mathrm{AEP} \sim \Delta \mathrm{ADB}$

(iv) In $\Delta \mathrm{PDC}$ and $\Delta \mathrm{BEC}$,

$\angle \mathrm{PCD}=\angle \mathrm{BCE}($ common $)$

$\angle \mathrm{PDC}=\angle \mathrm{BEC}=90^{\circ}$

$\therefore$ By AA similarity

$\Delta \mathrm{PDC} \sim \Delta \mathrm{BEC}$