In the adjoining figure, ABCD is a parallelogram.


In the adjoining figure, $A B C D$ is a parallelogram. If $P$ and $Q$ are points on $A D$ and $B C$ respectively such that $A P=\frac{1}{3} A D$ and $C Q=\frac{1}{3} B C$, prove that $A Q C P$ is a parallelogram.



We have:
B = ∠D                        [Opposite angles of parallelogram ABCD]
AD = BC and AB = DC      [Opposite sides of parallelogram ABCD]
Also, AD || BC and AB|| DC

It is given that $A P=\frac{1}{3} A D$ and $C Q=\frac{1}{3} B C$.

∴ ​AP = CQ                                   [∵ AD = BC]
In ∆​DPC and ∆​BQA, we have:

AB = CD, ∠B = ∠D and DP = QB        $\left[\because D P=\frac{2}{3} A D\right.$ and $\left.Q B=\frac{2}{3} B C\right]$

i.e., ∆​DPC ≅ ∆​BQA
∴​ PC  = QA

Thus, in quadrilatreal AQCP, we have:
  AP = CQ                   ...(i)
 PC  = QA                   ...(ii)
∴ ​AQCP is a parallelogram.

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