Answer: Option A. -> True
:
A
In the given figure,
AP = BP (Given)
AQ = BQ (Given)
PQ is common.
So, $\Delta$ PAQ $\cong$ $\Delta$ PBQ (SSS rule).
$\Rightarrow \angle$ APQ=$\angle$ BPQ (CPCT).
Also,AP = BP (Given) and PM is common
Hence, $△$ PMA $\cong$$△$PMB (SAS Rule)
So, AM = BM

Answer: Option A. -> perpendicular bisector
:
A

Consider the above figure.
Here, XY is the perpendicular bisector of a line AB.
Let P be any random point on XY.
In $△$ PMA $\cong$ $△$ PMB
AM = BM (Perpendicular bisector divides a line segment into two equal halves)
$\angle$ PMA =​$\angle$ PMB = 90${}^{\circ }$
Also PM is common side
So $△$ PMA $\cong$ $△$ PMB (SASRule)
$\therefore$ PA = PB (CPCT)
Hence, $△$ PAB is an isosceles triangle.
So, the given statement is true.

Answer: Option C. -> 70 degrees
:
C
All the standard angles and the angles which can be obtained by bisecting standard angles can be constructed just by using a ruler and compass. Here except ${70}^{\circ }$ all other angles can be constructed using ruler and compass.

Answer: Option B. -> Two parallel lines
:
B
Since two parallel lines do not intersect each other, anangle cannotbe formed between them. Thus in this case, angle bisection is not possible.

:
We need to draw just 2 arcs to draw a ${60}^{\circ }$ angle. As shown in the figure below, the first arc should have P as a centre. The second arc should have the same radius as the first and should have B as a centre.

Answer: Option B. -> Rectangle
:
B
In the case of a rectangle, the diagonal is not an angle bisector.
Consider a rectangleABCD.

Here, $\angle$ ADB =$\angle$ DBC (Alternate angles)
But,$\angle$ DBC is not equal to$\angle$ BDC (Angles opposite to unequal sides of a triangle are unequal)
$\therefore$$\angle$ ADB is not equal to$\angle$ BDC
So, diagonal DB does not bisect$\angle$ D.

Answer: Option C. -> Line segment
:
C
A perpendicular bisector can be drawn only if a figure has end points. Only a line segment has a definite length and hence it can be bisected by a perpendicular bisector.

Answer: Option C. -> Can't be determined
:
C
The information given above is insufficient to construct a triangle. So, the triangle cannot be constructed and its characteristics can’t be determined.

Answer: Option D. -> 10
:
D
In a triangle, the sum of thelengthsof any 2sidesof atrianglemust be greater than the thirdside. The third side can measure anything less than 11 units. Hence, the third side can be 10 units.

:
The procedure for Triangle Construction 2 is:
1. Draw the base BC of ∆ABC as given and construct ∠XBC of the required measure at B as shown.

2. From the ray, BX cut an arc equal to AB – AC at point P and join it to C as shown
3. Draw the perpendicular bisector of PC and let it intersect BX at point A as shown:

4. Join AC, ∆ABC is the required triangle.

Hence, it is clear that a perpendicular bisector is required in the process.