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Sagot :

Consider the triangle PAM and triangle PBM.

[tex]\begin{gathered} \angle PMA=\angle PMB\text{ (Each angle is right angle)} \\ AM=BM\text{ (M is perpendicular bisector of AB)} \\ PM\cong PM\text{ (Common side)} \\ \Delta\text{PMA}\cong\Delta\text{PMB (By SAS similarity)} \\ PA\cong PB\text{ (Corresponding part of Congurent triangle)} \end{gathered}[/tex]

Hence it is proved that,

[tex]PA\cong PB[/tex]

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