Question: PQRS is quadrilater PR and QS intersect at T . Prove that ar(PTQ)*ar(STR)=ar(PST)*ar(QTR).
I assume that ar(PTQ) means the area of the triangle PTQ.
The diagonals QS and PR intersect at T.
There are 4 angles at this point of intersection. They are both opposite and complementary.
Let the four angles be T, T', T and T', where T = 180 - T'.
From this we have sin(T) = sin(T').
The area of a triangle ABC is given by Area = (1/2)a.b.sin(C)
ar(PTQ) = (1/2)PT.QT.sin(T)
ar(STR) = (1/2)ST.RT.sin(T)
ar(PST) = (1/2)PT.ST.sin(T')
ar(QTR) = (1/2)QT.RT.sin(T')
ar(PTQ)*ar(STR) = (1/4)PT.QT.ST.RT.sin^2(T) = (1/4)PT.QT.RT.ST.sin^2(T)
ar(PST)*ar(QTR) = (1/4)PT.ST.QT.RT.sin^2(T') = (1/4)PT.QT.RT.ST.sin^2(T')
And sin(t) = sin(T'),
Hence ar(PTQ)*ar(STR) = ar(PST)*ar(QTR)