Given data:
The given figure.
In triangle PAM and PBM.
[tex]\begin{gathered} \angle PMA=\angle PMB=90^{\circ} \\ \bar{AM}\cong\text{ }\bar{\text{BM}}\text{ (Given)} \\ \bar{PM}\cong\bar{PM}\text{ (Common)} \\ \Delta PAM\cong\Delta PBM\text{ (Side-angle-side rule)} \\ PA\cong PB\text{ (Corresponding part of congruent triangle)} \end{gathered}[/tex]Thus, the PA is equal to PB.
