To determine:To prove:The congruency of /_PBC ~=/_PBD. Given information: The following information has been given /_BPC=/_BPD=90^(circ) /_C=/_D

To determine:To prove:The congruency of $\mathrm{\angle }PBC\stackrel{\sim }{=}\mathrm{\angle }PBD$.
Given information:
The following information has been given
$\mathrm{\angle }BPC=\mathrm{\angle }BPD={90}^{\circ }$
$\mathrm{\angle }C=\mathrm{\angle }D$
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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
$\mathrm{\angle }BPC=\mathrm{\angle }BPD={90}^{\circ }$
$\mathrm{\angle }C=\mathrm{\angle }D$
Thus, by virtue of similarity, third angle will also be congruent. Hence,
$\mathrm{\angle }PBC\stackrel{\sim }{=}\mathrm{\angle }PBD$
Thus, the given congruency is proved