★ To prove : –
ㅤㅤㅤ[tex]\boxed{\large\bf{\leadsto SA = NE}}[/tex]
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[tex]\begin{gathered}\rm{ \underline{ \underline{ \red{Given}}}}\begin{cases}\sf{SA || NE} & \\ \\\sf{Se || NA}&\end{cases}\\\\\end{gathered} [/tex]
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★ This means the given figure is a parallelogram .
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➜ In ∆ ESN and ∆ ASN :–
ㅤㅤㅤㅤSN = SN ㅤㅤㅤㅤ( common )
ㅤㅤㅤㅤ∠ ENS = ∠ NSAㅤ( alternate angles )
ㅤㅤㅤㅤ∠ SEN = ∠ NASㅤ( opposite angles in a parallelogram )
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[tex]\large\sf{\underline{\dag Hence,\:\triangle ESN \cong \triangle ASN ,\:by\: SAA\: criteria }}[/tex]
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★ Therefore , SA = NE ㅤㅤ ( by CPCT )
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★ Answer :–
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[tex]\large\rm{\longrightarrow \triangle ESN \cong \triangle ASN\: by \: reason \: SAA}[/tex]
