Question

ABCD ka program and BC is produced to Q such that ad equal to CD if PQ intersect BC point P such that triangle area Triangle BPC equal to area of triangle DPQ​

Answer 1

In ||gm ABCD ,

ar(△APC) = ar(△BCP) ---i)

[∵ triangles on the same base and between the same parallels have equal area]

Similarly, ar(△ADQ) = ar(△ADC) ---ii)

Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP)

ar(△DPQ) = ar(△ACP) ---iii)

From (i) and (iii) , we have

ar(△BCP) = ar(△DPQ)

or ar(△BPC) = ar(△DPQ)