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divide a line segment of length 9 cm internally in the ratio $4 : 3$. Also, give justification of the construction.
Given:
A line segment of length 9 cm.
To do:
We have to divide a line segment of length 9 cm internally in the ratio $4 : 3$.
Solution:
Steps of construction:
(i) Draw a line segment $AB = 9\ cm$.
(ii) Draw a ray $AX$ making an acute angle with $AB$.
(iii) From $B$, draw another ray $BY$ parallel to $AX$.
(iv) Cut off four equal parts from $AX$ and three parts from $BY$.
(v) Join $A_4$ and $B_3$ which intersects $AB$ at $P$.
$P$ is the required point which divides $AB$ in the ratio of $4 : 3$ internally.
Justification:
In $\triangle \mathrm{AA}_{4} \mathrm{P}$ and $\triangle \mathrm{BB}_{3} \mathrm{P}$,
$\angle A_{4} A P=\angle P B B_{3}$ ($\angle A B Y=\angle B A X$)
$\angle \mathrm{APA}_{4}=\angle \mathrm{BPB}_{3}$ (Vertically opposite angles)
Therefore, by AA similarity,
$\triangle \mathrm{AA}_{4} \mathrm{P} \sim \Delta \mathrm{BB}_{3} \mathrm{P}$
This implies,
$\frac{A A_{4}}{B B_{3}}=\frac{A P}{B P}$
$\frac{A P}{B P}=\frac{4}{3}$
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