Divide a line segment into three parts in the ratio $2: 3: 4$ in the same order.

(N/A) Data: $\overline{AB}$ is given.
To construct: $\overline{AB}$ is to be divided into three parts in the ratio $2: 3: 4$ from $A$.
Steps of construction:
$(1)$ In different half-planes of $\overleftrightarrow{AB}$ containing $\overline{AB}$,draw $\overrightarrow{AX}$ and $\overrightarrow{BY}$ such that $\angle XAB$ and $\angle YBA$ are congruent acute angles.
$(2)$ With some appropriate radius and centre $A$,draw an arc to intersect $\overrightarrow{AX}$ at $A_1$. Similarly,with centre $A_1$ and the same radius,draw an arc to intersect $\overrightarrow{AX}$ at $A_2$ such that $A-A_1-A_2$. Similarly,draw an arc with the same radius and centre $A_k$ to intersect $\overrightarrow{AX}$ at $A_{k+1}$ such that $A_{k-1}-A_k-A_{k+1}$,where $k=2, 3, 4, \dots, 8$. Thus,we obtain nine points $A_1, A_2, \dots, A_9$ on $\overrightarrow{AX}$ such that $AA_1 = A_1A_2 = \dots = A_8A_9$.
$(3)$ Now,with the same radius and beginning with point $B$ as centre,obtain nine points $B_1, B_2, \dots, B_9$ on $\overrightarrow{BY}$ such that $BB_1 = B_1B_2 = \dots = B_8B_9$.
$(4)$ Draw $\overline{AB_9}$,$\overline{A_2B_7}$,$\overline{A_5B_4}$,and $\overline{A_9B}$ so that $\overline{A_2B_7}$ intersects $\overline{AB}$ at $M$ and $\overline{A_5B_4}$ intersects $\overline{AB}$ at $N$.
Thus,we obtain points $M$ and $N$ which divide $\overline{AB}$ in the ratio $2: 3: 4$ from $A$,i.e.,$AM : MN : NB = 2: 3: 4$.
Justification: Here,the intercepts made on transversals $\overleftrightarrow{AX}$,$\overleftrightarrow{AB}$,and $\overleftrightarrow{BY}$ by $\overline{A_2B_7} \parallel \overline{A_5B_4}$ are proportionate.
Hence,$AM : MN : NB = AA_2 : A_2A_5 : A_5A_9 = B_9B_7 : B_7B_4 : B_4B = 2: 3: 4$.