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 triangle 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?

#### Solution

**Steps of construction**

**1.** Draw a line segment AB = 3 cm.

**2.** Now, draw a ray BY making an acute ∠ABY = 60°.

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

**4.** Again draw a ray AZ making an acute ∠ZAX’ = 60°. .....[∴ BY || AZ, ∴ ∠YBX’ = TAX’ = 60°]

**5.** With A as centre and radius equal to 5 cm draw an arc cut the point D on AZ.**6.** Now, join CD and finally make a parallelogram ABCD

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

**8.** From B 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.** From point C’ draw C’D’ || CD intersecting the extended line segment BD at D’. Then, AD’BC’ is the required triangle whose sides are of the corresponding sides of ΔDBC

**12.** Now draw a line segment D’A’ parallel to DA, where A’ lies on extended side BA i.e ray BX’.

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