Draw a parallelogram \[\mathbf{ABCD}\]in which \[\mathbf{BC}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{AB}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{cm}\]and angle \[\mathbf{ABC}\text{ }=\text{ }\mathbf{60}{}^\circ ,\]divide it into triangles \[\mathbf{BCD}\]and \[\mathbf{ABD}\]by the diagonal \[\mathbf{BD}.\]Construct the triangle \[\mathbf{BD}~\mathbf{C}~\]similar to triangle \[\mathbf{BDC}\]with scale factor\[\mathbf{4}/\mathbf{3}\]. Draw the line segment \[\mathbf{D}\mathbf{A}~\]parallel to \[\mathbf{DA}\]where \[\mathbf{A}\] lies on extended side\[\mathbf{BA}\]. Is \[\mathbf{A}\mathbf{BC}\mathbf{D}\] a parallelogram?

Steps of construction:

1. Define a boundary \[\mathbf{AB}=\mathbf{3}\text{ }\mathbf{cm}.\]
2. Draw a beam \[\mathbf{BY}\]making an intense \[\angle \mathbf{ABY}=\mathbf{60}{}^\circ .\]
3. With focus \[\mathbf{B}\]and span \[\mathbf{5}\text{ }\mathbf{cm},\]draw a circular segment cutting the point \[\mathbf{C}\text{ }\mathbf{on}\text{ }\mathbf{BY}.\]
4. Draw a beam \[\mathbf{AZ}\]making an intense \[\angle \mathbf{ZAX}’=\mathbf{60}{}^\circ .(\mathbf{BY}||\mathbf{AZ},\therefore \angle \mathbf{YBX}’=\mathbf{ZAX}’=\mathbf{60}{}^\circ )\]
5. With focus \[\mathbf{A}\]and range\[\mathbf{5}\text{ }\mathbf{cm}\], draw a circular segment cutting the point \[\mathbf{D}\text{ }\mathbf{on}\text{ }\mathbf{AZ}.\]
6. Join \[\mathbf{CD}\].
7. Subsequently we acquire a parallelogram \[\mathbf{ABCD}.\]
8. Join\[\mathbf{BD}\], the inclining of parallelogram \[\mathbf{ABCD}.\]
9. Draw a beam \[\mathbf{BX}\]downwards making an intense \[\angle \mathbf{CBX}.\]
10. Find \[{{B}_{1}},\text{ }{{B}_{2}},\text{ }{{B}_{3}},\text{ }{{B}_{4}}~on\text{ }BX\], with the end goal that\[B{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}={{B}_{3}}{{B}_{4}}.\]
11. Join \[{{B}_{4}}C\]and from\[{{B}_{3}}C\]

    define a boundary\[B4C||B3C\]

    converging the lengthy line portion \[\mathbf{BC}\text{ }\mathbf{at}\text{ }\mathbf{C}’.\]

12. Draw \[\mathbf{C}’\mathbf{D}’||\text{ }\mathbf{CD}\]crossing the drawn out line fragment \[\mathbf{BD}\text{ }\mathbf{at}\text{ }\mathbf{D}’.\]Then, at that point, ∆\[\mathbf{D}’\mathbf{BC}’\]is the necessary triangle whose sides are \[\mathbf{4}/\mathbf{3}\]of the comparing sides of ∆\[\mathbf{DBC}.\]
13. Presently define a boundary fragment \[\mathbf{D}’\mathbf{A}’||\text{ }\mathbf{DA},\]where \[\mathbf{A}’\] lies on the drawn out side \[\mathbf{BA}.\]
14. At last, we see that \[\mathbf{A}’\mathbf{BC}’\mathbf{D}’\] is a parallelogram where \[\mathbf{A}’\mathbf{D}’=\mathbf{6}.\mathbf{5}\text{ }\mathbf{cm}\text{ }\mathbf{A}’\mathbf{B}\text{ }=\text{ }\mathbf{4}\text{ }\mathbf{cm}\]and \[\angle \mathbf{A}’\mathbf{BD}’=\text{ }\mathbf{60}{}^\circ \] separation it into triangles \[\mathbf{BC}’\mathbf{D}’\text{ }\mathbf{and}\text{ }\mathbf{A}’\mathbf{BD}’\]by the slanting \[\mathbf{BD}’.\]