Question

In Fig., ABCD is a parallelogram. Points P and Q on BC trisects BC in three equal parts. Prove that ar (APQ) = ar (DPQ) = 1/6 ar(ABCD)

Answer 1

Through P and Q, draw PR and QS parallel to AB. Now PQRS is a parallelogram and its base PQ = 1/3 BC.

ar (APD) = 1/2 ar (ABCD) [Same base BC and BC || AD] (1) 

ar (AQD) = 1/2 ar (ABCD) (2) 

From (1) and (2), we get 

ar (APD) = ar (AQD) (3) 

Subtracting ar (AOD) from both sides, we get 

ar (APD) – ar (AOD) = ar (AQD) – ar (AOD) (4) 

ar (APO) = ar (OQD), 

Adding ar (OPQ) on both sides in (4), we get 

ar (APO) + ar (OPQ) = ar (OQD) + ar (OPQ) 

ar (APQ) = ar (DPQ) 

Since, ar (APQ) = 1/2 ar (PQRS), 

therefore ar (DPQ) = 1/2 ar (PQRS) 

Now, ar (PQRS) = 1/3 ar (ABCD) 

Therefore, ar (APQ) = ar (DPQ) 

= 1/2 ar (PQRS) = 1/2 × 1/3 ar (ABCD) 

= 1/6 ar (ABCD)