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°

Draw TE || AB.
\(\angle ABT+\angle ETB=180^0(AB||TE)\)
\(48^0+\angle ETB=180^0\)
\(\angle ETB=180^0-48^0=132^0\)
Similarly \(\angle CDT+\angle DTE=180^0\)
\(24^0+\angle DTE=180^0\)
\(\therefore \angle DTE=180^0-24^0\) = 156⁰
\(\therefore \angle BTE+\angle ETD=132^0+156^0\) = 288°
x = 288°

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°

In \(\triangle ABC\angle B=60^0\ and\ \angle C=70^0\)
\(\therefore\angle A=180^0-(60^0+70^0)\)
\(=180^0-130^0\) = 50⁰
In \(\triangle E=70^0\ and \angle D=60^0\)
\(\therefore \angle F=180^0-(70^0+60^0)\)
= 180° - 130° = 50°
\(\angle A=\angle F=50^0\)
\(\angle B=\angle D=60^0\)
\(\angle C=\angle E=70^0\)
By AAA congruency
\(\therefore\triangle ABC\cong\triangle FDE\)