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Question
In the given figure, PQRS is a ∥ gm. A straight line through P cuts SR at point T and QR produced at N. Prove that area of triangle QTR is equal to the area of triangle STN.
Solution
ΔPQT and parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.
∴ Δ(ΔPQT) = `(1)/(2)"A"`(parallelogram PQRS) ....(i)
ΔPSN and parallelogram PQRS are on the same base PS and between the same parallel lines PS and QN.
∴ Δ(ΔPSN) = `(1)/(2)"A"`(parallelogram PQRS) ....(ii)
Adding equations (i) and (ii), we get
∴ A(ΔPQT) + A(ΔPSN) = A(parallelogram PQRS)
⇒ A(quad. PSNQ) - A(ΔQTN) = A(parallelogram PQRS)
⇒ A(quad. PSNQ) - A(ΔQTN) = A(quad. PSNQ) - A(ΔSRN)
⇒ A(ΔQTN = A(ΔSRN)
Subtracting A(ΔRTN) from both the sides, we get
A(ΔQTN) - A(ΔRTN) = A(ΔSRn) - A(ΔRTN)
⇒ A(ΔQTR) = A(ΔSTN).
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