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I need help with a homework

User Chris Schmitz
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1 Answer

10 votes
10 votes

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.

User Andrew Janke
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