SSC CGL 23 Aug 2021 Shift 1 Paper

Given, $\angle \mathit{A}$ bisect by $\mathit{A}\mathit{D}$
$\mathit{M}=$ Mid-point
$\mathit{A}\mathit{B}=5\mathit{c}\mathit{m},\mathit{B}\mathit{C}=10\mathit{c}\mathit{m},\mathit{A}\mathit{C}=7.5\mathit{c}\mathit{m}$We know that,
$\frac{\mathit{B}\mathit{D}}{\mathit{D}\mathit{C}}=\frac{\mathit{A}\mathit{B}}{\mathit{A}\mathit{C}}$
$\frac{\mathit{B}\mathit{D}}{\mathit{D}\mathit{C}}=\frac{5}{7.5}=\frac{2}{3}$
$\mathit{B}\mathit{D}:\mathit{D}\mathit{C}=2:3$
$\mathit{B}\mathit{M}+\mathit{M}\mathit{C}=10\mathit{c}\mathit{m}$
where $\mathit{M}$ is mid-point.
So, $\mathit{B}\mathit{M}=\mathit{M}\mathit{C}=5\mathit{c}\mathit{m}$
When, $\frac{\mathit{B}\mathit{D}}{\mathit{D}\mathit{C}}$ cut in $\frac{2}{3}$
ratio of $10\mathit{c}\mathit{m}$ length.
That means $\mathit{B}\mathit{D}=\frac{2}{5}\times 10=4\mathit{c}\mathit{m}$
$\mathit{D}\mathit{C}=6\mathit{c}\mathit{m}$
We know that,
So, $\mathit{D}\mathit{C}-\mathit{M}\mathit{C}=5\mathit{c}\mathit{m}$
So, $\mathit{D}\mathit{C}-\mathit{M}\mathit{C}=\mathit{D}\mathit{M}$
$=6-5=1\mathit{c}\mathit{m}=\mathit{D}\mathit{M}$
So, the distance of $\mathit{D}$ from the mid-point of $\mathit{B}\mathit{C}$ is $1\mathit{c}\mathit{m}$.