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Answer 1

Given: In quad. PQRS, diagonals PR and QS intersect at T.

To prove: area(PTS) = area(RTQ)

Construction : From P and R, draw perpendiculars to QS, namely PE and RF

Proof : (a) In PET and RTF,

TP = TR (given)

PTE = RTF (vert. opp. angles)

PET = RFT = 90o

(AAS congruence)

area(PTE) = area(RTF) (congruent figures have equal area) ...(1)

And PE = RF (CPCT)

Now, in PSE and RQF,

PES = RFQ (right angle)

PE = RF (proved above)

PS = QR (given)

So, (RHS congruence)

Area (PSE) = area(RQF) ...(2)

Adding (1) and (2), we have:

Area(PTE) + area(PSE) = area(RTF) + area(RQF)

Or, area(PTS) = area(RTQ)

Answer 2

Given: In quad. PQRS, diagonals PR and QS intersect at T.

To prove: area(PTS) = area(RTQ)

Construction : From P and R, draw perpendiculars to QS, namely PE and RF

Proof : (a) In PET and RTF,

TP = TR (given)

PTE = RTF (vert. opp. angles)

PET = RFT = 90o

(AAS congruence)

area(PTE) = area(RTF) (congruent figures have equal area) ...(1)

And PE = RF (CPCT)

Now, in PSE and RQF,

PES = RFQ (right angle)

PE = RF (proved above)

PS = QR (given)

So, (RHS congruence)

Area (PSE) = area(RQF) ...(2)

Adding (1) and (2), we have:

Area(PTE) + area(PSE) = area(RTF) + area(RQF)

Or, area(PTS) = area(RTQ)