Chapter 7.2, Problem 3PSA

To determine

To prove : The congruency of $\angle PBC\cong \angle PBD$ .

Explanation of Solution

Given information : The following information has been given

  $\angle BPC=\angle BPD={90}^0$

  $\angle C=\angle D$

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

  $\angle BPC=\angle BPD={90}^0$

  $\angle C=\angle D$

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

  $\angle PBC\cong \angle PBD$

Thus, the given congruency is proved