Formula used : If any two triangles have two common angles, then by virtue of similarity, third angle will also be congruent

Proof : In triangles PBC and PBD, we have

\(\displaystyle\angle{B}{P}{C}=\angle{B}{P}{D}={90}^{{\circ}}\)

\(\displaystyle\angle{C}=\angle{D}\)

Thus, by virtue of similarity, third angle will also be congruent. Hence,

\(\displaystyle\angle{P}{B}{C}\stackrel{\sim}{=}\angle{P}{B}{D}\)

Thus, the given congruency is proved

Proof : In triangles PBC and PBD, we have

\(\displaystyle\angle{B}{P}{C}=\angle{B}{P}{D}={90}^{{\circ}}\)

\(\displaystyle\angle{C}=\angle{D}\)

Thus, by virtue of similarity, third angle will also be congruent. Hence,

\(\displaystyle\angle{P}{B}{C}\stackrel{\sim}{=}\angle{P}{B}{D}\)

Thus, the given congruency is proved