# 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?

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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