If area of 2similar triangles are …
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Posted by Vansh Walia 4 years, 10 months ago
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Sia ? 4 years, 10 months 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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