Question

Divide a line segment $6cm$ in the ratio $4:3$. Prove your assertion.

Answer 1

Steps of construction
1. Draw a line segment $AB=6cm$.
2. Draw a ray $AX$, making an acute angle $\angle BAX$.
3. Along $AX$, mark $(4+3)=7$ points $A_1,A_2,A_3,A_4,A_5,A_6$ and $A_7$ such that $A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4=A_4{A}_5=A_5{A}_6=A_6{A}_7$.
4. Join $A_{7}B$.
5. From $A_4$, draw $A_{4}C\parallel A_{7}B$, meeting $AB$ at $C$. Then, $C$ is the point on $AB$, which divides it in the ratio $4:3$.
Thus, $AC:CB=4:3$
Proof : Let $A{A}_1=A_1{A}_2=...=A_6{A}_7=x$.In $△AB{A}_7$, we have $A_{4}C\parallel A_{7}B$.$\therefore \frac{AC}{CB}=\frac{A{A}_4}{A_4{A}_7}=\frac{4x}{3x}=\frac{4}{3}$ [By Thales' theorem].

Hence, $AC:CB=4:3$.

Alternative method
Steps of construction
1. Draw a line segment $AB=6cm$.
2. Draw a ray $AX$, making an acute angle $\angle BAX$.
3. Draw a ray $BY$ parallel to $AX$ by making $\angle ABY=\angle BAX$.
4. Locate the points $A_1,A_2,A_3,A_4$ on $AX$ and $B_1,B_2,B_3$ on $BY$ such that $A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4=B{B}_1=B_1{B}_2=B_2{B}_3$.
5. Join $A_4{B}_3$, intersecting $AB$ at a point $C$. Then, $AC:CB=4:3$.
Proof : Here, $A{A}_4\parallel B{B}_3\Rightarrow \angle CA{A}_4=\angle CB{B}_3$.
$\therefore \angle CA{A}_4=\angle CB{B}_3$ (alt. int. $\angle s$)
$\angle AC{A}_4=\angle BC{B}_3$ (vert. opp. $\angle s$)
$\angle A{A}_{4}C=\angle B{B}_{3}C$ (alt. int. $\angle s$)
$\therefore △CA{A}_4$ is similar to $△CB{B}_3$
$\Rightarrow \frac{AC}{CB}=\frac{A{A}_4}{B{B}_3}=\frac{4}{3}$
Hence, $AC:CB=4:3$