ABCD is a parallelogram. Points P …

Ans.

Given: ABCD is a parallelogram in which points P and Q trisects the BC.

\(=> BP = PQ = QC  = {1\over 3}BC\)

To prove: \(ar (\triangle APQ) = ar (\triangle DPQ) = {1\over 6} ar(\|gm \space ABCD)\)

Construction : Through P and Q, draw PR and QS parallel to AB. Now PQRS is a parallelogram and its base \( PQ = {1\over 3} BC\)

Proof : 

Now, \(\triangle APD \space and \space \triangle AQD\) lie on the same base AD and between the same parallel AD and BC.
\(ar (\triangle APD) \space = ar \space( \triangle AQD)\)        .......... (1)

Subtracting \(ar (\triangle AOD)\) from both sides, we get

\(=> ar (\triangle APD) - ar (\triangle AOD) \space = ar \space( \triangle AQD) -ar (\triangle AOD) \)

\(=> ar (\triangle APO) = ar (\triangle OQD) \)      ............... (2)

Adding \(ar (\triangle OPQ)\) on both sides in (2), we get

\(=> ar (\triangle APO) + ar (\triangle OPQ) = ar (\triangle OQD) + ar (\triangle OPQ)\)

\(=> ar (\triangle APQ) = ar (\triangle DPQ)\)    ....... (3)

Again, triangle APQ and parallelogram PQSR are on the same base PQ and between same parallels PQ and AD

\(ar (\triangle APQ) ={1\over2} ar (\|gm \space PQRS)\)   ........ (4)

Now, 

\({ar(\| gm \space ABCD )\over ar(\| gm \space PQRS )} = {BC \times height\over PQ\times height } = {3PQ\over PQ} = 3\)

\(=> ar(\| gm \space PQRS ) = {1\over 3} ar(\| gm \space ABCD ) \)     ....... (5)

\(=> ar (\triangle APQ) ={1\over2} \times {1\over 3}ar (\|gm \space ABCD)\)

\(=> ar (\triangle APQ) ={1\over6}ar (\|gm \space ABCD)\)   ....... (6)

From (3) and (6)

\(=> ar (\triangle APQ) = ar (\triangle DPQ)={1\over6}ar (\|gm \space ABCD)\)

Hence Proved