Question

I need help with a homework

Answer 1

Given data:

The given figure.

In triangle PAM and PBM.

`\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}`

Thus, the PA is equal to PB.