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
AB = AC (Given)
BM = MC (AM is the median of the \(\triangle\)ABC)
AM is common (By SSS congruency)
\(\therefore\triangle ABM\cong\triangle ACM\)

In the given figure AC and BD bisect each other at O.
\(\therefore\) OA = OC (Given) ; OB = OD (Given)
\(\angle AOB=\angle COD\) (vertically opposite angles)
By SAS congruency
\(\triangle AOB\cong\triangle OCD\)

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 \(\triangle PXY\ and \triangle PXZ\)
PX is common.
XY = XZ (Given)
PY = PZ (Given)
By SSS congruency
\(\triangle PXY\cong\triangle PXZ\)

In \(\triangle ABD\ and\ \triangle CDB\)
AB = CD (Given)
AD = BC (Given)
BD is common
By SSS congruency
\(\triangle ABD\cong\triangle CDB\)

In \(\triangle PQR\ and \ \triangle ABC\)
PQ = AB (Given)
RQ = BC (Given)
\(\therefore \triangle ABC\) is congurent to \(\triangle PQR\)

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°

\(\angle 1:\angle 2=11:7\)
\(\angle 1+\angle 2=180^0\) (Angle of a straight line)
Divide 180° in the ratio 11 : 7
Total ratio 11 + 7 = 18
\(\angle 1=\frac{11}{18}\times180^0=110^0\)
\(\angle 2=\frac{7}{18}\times180^0=70^0\)
\(\angle 3=\angle 1=110^0\)(Vertically opposite angles)
\(\angle 4=\angle 2=70^0\)(Vertically opposite angles)
\(\angle 5=\angle 1=110^0\)(Corresponding angles)
\(\angle 6=\angle 2=70^0\)(Corresponding angles)
\(\angle 7=\angle 5=110^0\)(Vertically opposite angles)
\(\angle 8=\angle 6=70^0\)(Vertically  opposite angles )
\(\therefore\) ∠1 = 110°, ∠2 = 70°, ∠3 = 110°, ∠4 = 70°, ∠5 = 110°, ∠6 = 70°, ∠ 7 = 110° ∠8 = 70°.