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Sia 🤖 9 months, 3 weeks ago

Given:

{tex}∆ABC\sim∆PQR{/tex}

and {tex}ar∆ABC=ar∆PQR{/tex}

To prove: {tex}∆ABC\cong∆PQR{/tex}

Proof: {tex}∆ABC\sim∆PQR{/tex}

Also {tex}\operatorname { ar } ( \Delta A B C ) = \operatorname { ar } ( \Delta P Q R ){/tex} (given)

or, {tex}\frac { \operatorname { ar } ( \Delta A B C ) } { \operatorname { ar } ( \Delta P Q R ) } = 1{/tex}

Or {tex}\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{CA^2}{RP^2}=1{/tex}

Or {tex}\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}=1{/tex}

Hence we get that

AB=PQ,BC=QR and CA=RP

Hence {tex}∆ABC\cong ∆PQR{/tex}

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