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Question
ΔPQR and ΔSQR are on the same base QR with P and S on opposite sides of line QR, such that area of ΔPQR is equal to the area of ΔSQR. Show that QR bisects PS.
Solution
Join PS. Suppose PS and QR intersect at O. Draw PM and SN perpendicular to QR.
ar(ΔPQR) = ar(ΔSQR)
Thus ΔPQR and ΔSQR are on the same base QR and have equal area.
Therefore, their corresponding altitudes are equal i.e. PM = SN.
Now,
In ΔPMO and ΔSNO.
∠1 = ∠2 ...(vertically opposite angles)
∠PMO = ∠SNO ...(right angles)
PM = SN
Therefore, ΔPMO ≅ ΔSNO ..(AAS axiom)
⇒ PO = OS
⇒ QR bisects PS.
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