It is a quadrilateral

It is not a parallelogram

\(\angle XOZ+\angle OZY=180^0\) (angle of a straight line)
(x° - 20°) + 60° = 180°
x°+40° = 180°
x° = 180°- 40°
= 140°
The value of x = 140°

Complement of 70° = 90° - 70°
=20°

In \(\triangle ABC\ and\ \triangle EDF\)
AB = EF (Given)
\(\angle ABC=\angle EDF\) (Given)
\(\angle ACB=\angle EFD\) (Given)
By ASA congruency
\(\triangle ABC\cong\triangle EDF\)

In the given figure BD bisect AC
In \(\triangle AOB\ and \triangle OCD\)
OA = OC (Given)
\(\angle AOB=\angle DOC\) (vertically opposite angles)
\(\angle B=\angle D\) (Given)
By ASA congruency \(\triangle AOB\cong\triangle OCD\)

In the given figure AB II CD, AD is the transversal
\(\angle CDA=\angle BAD\)
= 53° (alternate angles are equal)
In \(\triangle ECD,\angle D=\angle A=53^0\) (Alternate angles are equal)
\(\angle E+\angle C+\angle D=180^0\) (sum of the angles of a triangle)
x° + 38° + 53° = 180°
x° = 180°- 91°​​​​​​​ = 89°​​​​​​​
x = 89°

Through T draw TE|| AB.
\(\therefore \angle BAT+\angle ATE=180^0(AB||TE)\)
\(140^0+\angle ATE=180^0\)
\(\angle ATE=180^0-140^0=40^0\)
Similarly \(\angle ETC+\angle TCD=180^0(TE||CD)\)
\(\angle ETC+150^0=180^0\)
\(\angle ETC=180^0-150^0=30^0\)
\(x=\angle ATE+\angle ETC\)
= 40° + 30° = 70°
x = 70°

Steps for construction:
Step 1: Draw the ΔPQR with the given measurements.
Step 2: Construct altitudes from any two vertices P and Q to their opposite sides QR and PR respectively.
Step 3: The point of intersection of the altitude H is the orthocentre of the given  ΔPQR.