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.

Answer 1

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.