Question

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

Answer 1

Step 1. Draw a line $AB=3\mathrm{cm}$ and then draw a ray $BY$ making an acute $\angle ABY=60°$.

Step 2. Considering $B$ as center and radius $5\mathrm{cm}$, draw an arc cutting the point $C$ on $BY$and then draw a ray $AZ$ making an acute $\angle ZAX\text{'}=60°\left[BY\left|\right|AZ,\therefore \angle YBX\text{'}=ZAX\text{'}=60°\right]$

Step 3. With centre $A$ and radius $5\mathrm{cm}$, draw an arc cutting the point $D$ on $AZ$ and then Join $CD$

Thus parallelogram $ABCD$ is constructed.

Step 8. Join $BD$, the diagonal of parallelogram $ABCD$.

Step 9. Draw a ray $BX$ downwards making an acute $\angle CBX$.

Step 10. Locate four points $B_1,B_2,B_3,B_4$ on $BX$, such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$

Step 11. Join $B_{4}C\text{'}$ and from $B_{3}C$ draw a line $B_{4}C\text{'}\left|\right|B_{3}C$ intersecting the extended line segment $BC$ at $C\text{'}$.

Step 12. Draw $C\text{'}D\text{'}\left|\right|CD$ intersecting the extended line segment $BD$ at $D\text{'}$. Then, $∆D\text{'}BC\text{'}$is the required triangle whose sides are $\frac{4}{3}$ of the corresponding sides of $∆BDC$.

Step 13. Now draw a line segment $D\text{'}A\text{'}\left|\right|DA$, where $A\text{'}$ lies on the extended side $BA$.

Step 14. Finally, we observe that$A\text{'}BC\text{'}D\text{'}$ a parallelogram in which in which $BC=5\mathrm{cm},AB=3\mathrm{cm}$and angle $ABC=60°$, divide it into triangles $BCD$ and $ABD$ by the diagonal $BD$.