Question

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$\\ \\ \boxed{\begin{array}{c}\sf Prove\:that \\ \tt{ PA= PB} \\ {\boxed{\bf Diagram\: attached}}\end{array}}$

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Answer 1

$\bf{Given : } \rm{}$

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$\bf{To \: Prove : } \: \rm{PA = PB}$

$\bf{Proof :}$

$\: \: \: \: \: \: \: \: \: \: \sf∠BPA = 120⁰$

$\sf{Also, ∠OPA = 60⁰. }\\ \sf{So, ∠BPO = 120⁰ - 60⁰ = 60⁰}$

$\textsf{In ΔPOB and ΔPOA,} \\ \\\sf{∠OPA = ∠BPO = 60⁰ \: \: [Given]} \\ \sf{∠OAP = ∠OBP = 90⁰ \: \: [Given]}$

$\\ \sf{As, ∠OPA + ∠PAO + ∠AOP =∠OPB + ∠PBO + ∠BOP}$

$\textsf{[Angle Sum Property of triangle]}$

$\sf{ \cancel{∠OPA} + \cancel{∠PAO} + ∠AOP = \cancel{∠OPB} + \cancel{∠PBO} + ∠BOP}$

$\sf So, ∠BOP = ∠AOP$

$\sf Now, in \: ΔPOB \: and \: ΔPOA, \\ \sf \longmapsto∠OPA = ∠BPO = 60⁰ \: \: [Given] \\ \sf \longmapsto OP = OP \: \: [Common] \\ \sf \longmapsto ∠BOP =∠AOP \: \: [Proved \: above]$

$\sf \large \underline\purple{ Hence, ΔPOB \cong ΔPOA } \\ \sf \large \underline\orange{ by \: ASA \: congruency \: axiom}$

$\huge \sf{\pink{\therefore \: by \: CPCT, PA = PB }}\green{\checkmark}$