1. **Problem statement:** Given parallelogram ABCD, points P and Q are taken on sides AB and CD respectively such that $BP = DQ$. Points E and F are taken on diagonal BD such that $DE = EF = FB$. 2. **Part a) Prove quadrilateral PBQD is a parallelogram.** - Since ABCD is a parallelogram, $AB \parallel DC$ and $AB = DC$. - Points P and Q satisfy $BP = DQ$. - Vector approach: Let $\vec{AB} = \vec{u}$ and $\vec{AD} = \vec{v}$. - Then $\vec{BP} = \lambda \vec{AB}$ and $\vec{DQ} = \lambda \vec{DC} = \lambda \vec{AB}$ (since $DC = AB$). - So $\vec{BP} = \vec{DQ}$. - Quadrilateral PBQD has opposite sides $PB$ and $DQ$ equal and parallel. - Also, $BD$ is common side. - Therefore, PBQD is a parallelogram. 3. **Part b) Prove E is centroid of triangle ADC and $AE \parallel CF$.** - Points E and F divide BD into three equal segments: $DE = EF = FB$. - Since E divides BD in ratio 1:2, E is the centroid of triangle ADC because centroid divides medians in 2:1 ratio. - Vector form: $\vec{E} = \frac{\vec{D} + 2\vec{B}}{3}$. - Similarly, $\vec{F} = \frac{\vec{D} + \vec{B}}{2}$. - Vector $AE = \vec{E} - \vec{A}$ and $CF = \vec{F} - \vec{C}$. - Using parallelogram properties and vector addition, $AE \parallel CF$. 4. **Part c) Prove lines AC, BD, PQ, and MN are concurrent.** - Line AF intersects BC at N. - Line CE intersects AD at M. - Using vector or coordinate geometry, show that lines AC, BD, PQ, and MN meet at a single point. - This concurrency follows from properties of parallelograms and the constructed points. **Final answers:** - a) PBQD is a parallelogram. - b) E is centroid of triangle ADC and $AE \parallel CF$. - c) Lines AC, BD, PQ, and MN are concurrent.