As no angles are given, we need to know all three sides to construct a triangle. Thus, triangle ADC and ABD cannot be constructed as first step because side AD’s length is not known.

A quadrilateral is made up of two triangles. Diagonal is the common side for both the triangles. If all the four sides and a diagonal is known, then we can construct a quadrilateral by constructing two triangles. The two triangles are constructed using SSS property.

The line segment for which both the angles are given should be drawn first. For e.g. in the case of line segment MI, both angles M and I are known. Similarly, for IS, the angle I and S are known. Thus, either MI or IS can be drawn, because angles at both end points of both line segments are known.

Sum of the two sides of the triangle is greater than 3rd side
let, 1st side be $a=2$, 2nd be $b=3$, 3rd be $c$.
As per above condition, 
$a+b>c$ $\Rightarrow$ $2+3>c$ $\Rightarrow$ $c<5$
$a+c>b$ $\Rightarrow$ $2+c>3$ $\Rightarrow$ $c>1$
so, $1<c<5units$

If I combine 2 equilateral triangles, I get a quadrilateral with all sides equal. Since the angle of an equilateral triangle is ${60}^{\circ }$, no sides will be perpendicular. The quadrilateral formed cannot be a square. Hence, it is a rhombus.

Consider quadrilateral ABCD above. The triangles that can be constructed with diagonals are the following- ABC, DBC, BAD, CAD. Hence, 4 triangles are possible.

We can't draw a unique quadrilateral with all the measurements which are given. This is because we need the measurement of two adjacent sides and 3 angles to construct the quadrilateral ABCD. But with BC, we will have 6 sets of measurements and two known adjacent sides. Hence, we will be able to construct the quadrilateral ABCD.

If we have the measures of all the sides of a quadrilateral and its one diagonal, we can draw two triangles and hence, form a quadrilateral.

ABCD comprises of two triangle $△$ ADB & $△$ CDB

In Δ ADB, AB = 3cm, BD = 8cm, and AD = 4cm

Since, AB + AD  $<$ BD

$△$ ADB  can't be constructed.

So, quadrilateral ABCD can't be constructed.

Measurement of all the sides or 2 angles and 1 side or 1 angle and 2 sides are required to construct a unique triangle. Using 3 angles, we can construct many triangles, not a unique triangle. For e.g. there are infinite equilateral triangles though they have all the angles equal to ${60}^{\circ }$.

Step1: Draw rough diagram of the parallelogram ABCD

Here, $\angle$ DAC = $\angle$  ACB= ${15}^{\circ }$ - (1) (Alternate interior angles)
In $△$ ABC,
$\angle$ A + $\angle$ B + $\angle$ C must be ${180}^{\circ }$
$\angle$ A + $\angle$ B + $\angle$ C = ${25}^{\circ }$ + ${110}^{\circ }$ + ${15}^{\circ }$             (from (1))
$\angle$ A + $\angle$ B + $\angle$ C = ${150}^{\circ }$
The sum of angles in a triangle should always be equal to ${180}^{\circ }$. So, this construction is not possible.