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

Answers (1)

Solution

Steps of construction
1.   Draw, A line AB = 3 cm
2  Draw a ray by making  \angle ABP= 60^{\circ}
3.   Taking  B as centre and radius equal to 5 cm. Draw an arc which cut BP at point C
4.   Again draw ray AX making  \angle {Q}'AX= 60^{\circ}
5.   With A as centre and radius equal to 5 cm draw an arc which cut AX at point D
6.   Join C and D Here ABCD is a parallelogram           
7.   Join BD , BD is a diagonal of parallelogram ABCD
8.   From B draw a ray BQ with any acute angle at point B i.e., \angle CBQ  is acute angle
9.   Locate 4 points B_{1},B_{2},B_{3},B_{4}  on BQ with equal distance.
10. Join B_{3}C  and from  B_{4},{C}'  parallel to B_{3}C  which intersect at point {C}'
 11. From point  {C}' draw line {C}'{D}'  which is parallel to CD
 12. Now draw a line segment {D}'{A}'  parallel to DA
 Note : Here {A}',{C}'  and {D}'  are the extended sides.
13.  {A}'B{C}'{D}' is a parallelogram in which {A}'{D}'= 6\cdot 5\, cm  and {A}'{B}= 4\, cm  and < {A}'B{D}'= 60^{\circ} divide it into triangles B{C}'{D}'  and {A}'{BD}'  by the diagonal {BD}'

           

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