Question

∆ABC~∆PQR,ADis perpendicular to BC , PS is perpendicular to QR then prove A(∆ABC)/A(∆PQR)=BC²/QR²​

Answer 1

Step-by-step explanation:

Given:ΔABC~ΔPQR

Ar(ABC)=Ar(PQR)

To prove :ΔABC is congruent to ΔPQR

Proof :ΔABC~ΔPQR(given)

Ar(ABC)/Ar(PQR)=AB²/PQ²=BC²/QR²=AC²/PR²

Ar(ABC)/Ar(ABC)=AB²/PQ²=BC²/QR²=AC²/PR²

1=AB²/PQ²=BC²/QR²=AC²/PR²

1=AB²/PQ², 1=BC²/QR², 1=AC²/PR²

AB²=PQ², BC²=QR², AC²=PR²

AB=PQ, BC=QR,AC=PR

Now in ΔABC and ΔPQR

AB=PQ

BC=QR

AC=PR

so by SSS congruency ΔABC is congruent to ΔPQR

Hence, proved

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