In the figure, in $\triangle MNL$ and $\triangle PQR$, $\angle M = \angle Q = 70^\circ$, $MN = 3$ cm, $ML = 4.5$ cm, $PQ = 2$ cm, and $QR = 3$ cm. Then, the following correct relation will be: 
Step 1: Given data.
For $\triangle MNL$: $MN = 3$ cm, $ML = 4.5$ cm, and $\angle M = 70^\circ$.
For $\triangle PQR$: $PQ = 2$ cm, $QR = 3$ cm, and $\angle Q = 70^\circ$.
Step 2: Compare sides including the equal angles.
\[ \dfrac{MN}{QR} = \dfrac{3}{3} = 1, \quad \dfrac{ML}{PQ} = \dfrac{4.5}{2} = 2.25 \] These are not equal, but let’s check other possible corresponding sides. If we consider $\triangle NML$ and $\triangle QPR$: \[ \dfrac{MN}{QP} = \dfrac{3}{2} = 1.5, \quad \dfrac{ML}{QR} = \dfrac{4.5}{3} = 1.5 \] The sides are in the same ratio and included angles are equal (\(70^\circ\)).
Step 3: Conclusion.
Hence, by the SAS similarity criterion, \[ \triangle NML \sim \triangle QPR \]
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).