Answer: Option B. -> Drawing an obtuse angle would make my constructions very congested.
:
B
Suppose we draw $PM$ such that the angle $\angle QPM$ is an obtuse angle. Next we mark $(m+n)$ arcs on the ray PM such that $P{P}_1$ = $P_1{P}_2$ = $P_2{P}_3$ = … = $P_{m+n-1}$$P_{m+n}$.
For our convenience let us assume $m+n=5$ so we can explain things much easier. (In our case the required ratio could be 1:4 or 2:3 or 3:2 or 4:1 etc...)
Now in the $△$$QP{P}_5$,
If, $\angle$$QP{P}_5$ is obtuse then $\angle$$Q{P}_{5}P$ and $\angle$$PQ{P}_5$ would become very small (compared to the case where $\angle$ $QP{P}_5$ is acute) and it would be difficult to draw a line parallel to $Q{P}_5$ (Because small angles and hence small radii of arcs are involved in construction of parallel lines)
This will make drawing of parallel lines very congested compared to drawing parallel lines on line segment PN with acute angle $\angle QPN$.

You can see from the diagram that $P_4{Q}^{\prime }\parallel P_{5}Q$ is more closely spaced compared to$P_4^{\prime }{Q}^{\prime }\parallel P_5^{\prime }Q$.
So we prefer to draw an acute angle to make it more clear and spacious i.e. to avoid congestion.

Answer: Option A. -> 34
:
A
As we draw $\Delta A^{\prime }{C}^{\prime }B$ similar to$\Delta ACB$ with corresponding side ratio is $\frac{3}{4}$:
$\frac{B{A}^{\prime }}{BA}=\frac{B{C}^{\prime }}{BC}=\frac{A^{\prime }{C}^{\prime }}{AC}=\frac{3}{4}$

Answer: Option A. -> Corresponding angles are equal
:
A

Here we use the principle that when corresponding angles are equal, the lines are parallel.
$\angle A^{\prime }{C}^{\prime }B=\angle ACB\to$ AC||A'C'.

Answer: Option B. -> 4:5
:
B

According to the construction, $\Delta B{B}_{4}A\sim \Delta B{B}_5{A}^{\prime }$
And for similar triangles, the ratio of corresponding sides is $\frac{AB}{A^{\prime }B}$.
The ratio of corresponding sides is $4:5$.

Answer: Option B. -> bisect
:
B
If we draw two arcs from any random points A and B with radius larger than half the distance between AB, the perpendicular bisector of AB passes through the point of intersection of the two arcs. If you consider the line AB to be a chord then the perpendicular of AB passing through centre will bisect the chord. By combining the two concepts we can arrive at the conclusion that the perpendicular passing though P and O bisects the arc AB. Therefore, OP will bisect the $\angle AOB$.

Answer: Option B. -> 1, 4, 2, 3
:
B
Steps to be followed to for the construction of a similar triangle of $\Delta ABC$ is given below.
1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2.Locate 4(the greater of 3 and4 in$\frac{3}{4}$) points $B_1,B_2,B_3\text{and}{B}_4$ on BX so that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$
3.Join $B_{4}C$ and raw a line through $B_3$( the third point, 3 being smaller of 4 in$\frac{3}{4}$) parallel to$B_{4}C$ to intersect BC at C'.
4.Draw a line through C' parallel to the line CA to intersect BA at A'.

Answer: Option A. -> Corresponding angles are equal
:
A

Here we use the principle that when corresponding angles are equal, the lines are parallel.
$\angle A{A}_{3}C=\angle A{A}_{5}C$ as $A_{3}C$ is parallel to $A_{5}B$

Answer: Option A. -> acute angle
:
A
For the construction of a similar triangle of $\Delta ABC$, we draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

Answer: Option C. -> m+n
:
C

The number of arcs to be drawn, when a single ray is used to divide a line segment in the ratio $m:n$, where $m$ and $n$ are co-prime, is $m+n$. 

Answer: Option D. -> $17:5$
:
D

Given $\frac{AC}{BC}=\frac{5}{12}$

Therefore, $\frac{BC}{AC}=\frac{12}{5}$

$\frac{AB}{AC}=\frac{AC+BC}{AC}=1+\frac{BC}{AC}=1+\frac{12}{5}=\frac{17}{5}$

So, the required ratio $=17:5$