Draw a parallelogram ABCD in which BC=5 cm, AB=3 cm, and ∠ABC=60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangles BD’C’ similar to ∆BDC with scale factor 4/3. Draw the line segment D’A’ parallel to DA, where A’ lies on extended side BA. Is A’BC’D’ a parallelogram?

Steps of construction

1. Draw a line AB=3 cm.

2. Draw a ray BY making an acute ∠ABY=60°.

3. With B as center and radius equal to 5 cm draw an arc cut the point C on BY.

4. Draw a ray AZ making an acute ∠ZAX’=60°.(BY││AZ, ∴ ∠YBX’=ZAX’=60°)

5. With A as center and radius equal to 5 cm draw an arc cut the point D on AZ.

6. Join CD and finally make a parallelogram ABCD.

7. Join BD, which is a diagonal of parallelogram ABCD.

8. Draw any ray BX downwards making an acute ∠CBX.

9. Locate 4 points B_{1},B_{2},B_{3},B_{4} on BX, such that BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}.

10. Join B_{4}C and from B_{3}C draw a line B_{4}C’||B_{3}C intersecting the extended line segment BC at C’.

11. Draw C’D’CD intersecting the extended line segment BD at D’. Then, ∆D’BC’ is the required triangle whose sides are 4/3 of the corresponding sides of ∆DBC.

12. Now draw a line segment D’A’DA, where A’ lies on the extended side BA

13. Finally, we observe that A’BC’D’ is a parallelogram in which A’D’=6.5 cm A’B = 4 cm and ∠A’BD’= 60° divide it into triangles BC’D’ and A’BD’ by the diagonal BD’.

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