\(23^o+40^o+35^o+\angle D=360^o\) (Angle sum property of a quadrilateral)
\(\angle D=262^o\)
\(x^o=Reflex\angle D=360^o-262^0\)
=98^o

Join AC,

\(In\ \triangle ABC\ Ext\angle 4=\angle ACB+\angle CAB...(i)\)
Again, in \(\triangle ACD\ Ext.\angle 3=\angle DAC+\angle DCA...(ii)\)
Adding (i) and (i), we get
\(\angle 3+\angle 4=(\angle ACB+\angle DCA)+(\angle CAB+\angle DAC)\)
\(=\angle 1+\angle 2\)
\(\therefore \angle 3+\angle 4=\angle 1+\angle 2\)

y=180^o-(25^o+125^o)(Linear pair)
\(\Rightarrow\) y=30^o
y+z=110^o (Exterior angle)
z=110^o-30^o=80^o
x+25^o=z (Exterior angle)
x=80^o-25^o=55^o

In \(\triangle ABC,\angle A+\angle B+\angle C=180^o\)(Angle sum property)
\(\Rightarrow 90^o+\angle B+\angle C=180^o(\angle A=90^o)\)
\(\Rightarrow \angle B+\angle C=90^o\)
\(\therefore \angle C=\angle ACB=90^o-\angle B...(i)\)
Also in \(\triangle ALB,\)
\(\angle ALB+\angle BAL+\angle B=180^o\)  (Angle sum property)
\(\Rightarrow 90^o+\angle BAL+\angle B=180^o\)
\(\angle BAL=90^o-\angle B....(ii)\)
Hence, from (i) and (ii), we get
\(\angle BAL=\angle ACB\)

EG is the bisector of \(\angle AEF\)
\(\angle AEG=\angle GEF=a\)
Similarly, \(\angle EFH=\angle HFD=b\)
\(\angle GEF=\angle EFH(\because a=b)\)
But these are alternate interior angles
EG||FH
Again, \(\angle AEF=2a\)
and \(\angle EFD=2b\)
\(\angle AEF=\angle EFD=2a \ or\ 2b\)
But these are alternate angles 
AB||CD.

\(\angle BMN+\angle DNM=180^o\) (Interior angles)
\(\Rightarrow \frac{\angle BMN}{2}+ \frac{\angle DMN}{2}=\frac{180^o}{2}\Rightarrow\angle1+\angle2=90^o\)
\(\angle 1+\angle2+\angle3=180^o\Rightarrow \angle3=90^o\)
PM and PN intersect at right angle.

Draw CF||AB, produce ED to H

AB||CD||CF
\(\angle 1+\angle BCF=180^o...(i)\)
\(\angle 3=180^o-\angle4\)
\(\Rightarrow \angle4=180^o-\angle3..(ii)\)
Again CF||EH
\(\angle DCF+\angle 4=180^o...(iii)\)
Adding (i) and (iii), we get
\(\angle1+\angle 4+\angle BCF+\angle DCF=180^o+180^o=360^o\)\(​​​​​​​\angle1+\angle2+\angle4=360^o\)
Using (ii) we get
\(\angle1+\angle2+180^o-\angle3=360^o\)
\(\Rightarrow \angle1+\angle2-\angle3=360^o-180^o=180^o\)

Given PQ||RS
Given \(\angle 1=\angle 2\ and \angle 3 =\angle 4\)
But, \(\angle 1=\angle 3 \) (Corres angles)
\(2\angle 1=2\angle3\)
\(\angle MPB=\angle PRD\)
Hence, AB||CD.

y=180^o-(30^o+20^o)=130^o
l||m \(\Rightarrow\)x+100^o=180^o\(\Rightarrow\)x=80^o
x+y=210^o,y-x=50^o
(x+y):(y-x)=21:5

x+y+w+z=360^o
\(\Rightarrow\) 2(x+y)=360^o \((\because x+y=w+z)\)
\(\Rightarrow\) x+y=180^o
\(\therefore\) AOB is a straight line.

Two lines AB and CD intersect at point O. Also, OM and ON are t6he bisectors of \(\angle AOC\) and \(\angle BOD\) respectively.

To prove: MON is a straight line/
Prove: Since the sum of all the angles around a point O is 360^o , we have
\(\angle AOC+\angle BOC+\angle BOD+\angle AOD=360^o\)
\(\Rightarrow 2\angle MOC+2\angle BOC+2\angle BON=360^o\)
\([\because \angle BOC=\angle AOD\)(ver:opp, \(\angle S\) OM is bisector of \(\angle AOC.ON\) is bisector of \(\angle BOD\)]
\(\Rightarrow \angle MOC+\angle BOC+\angle BON\)
=180^o
\(\Rightarrow \angle MON=180^o\)
\(\therefore \angle MON\) is a straight angle
Hence, MON is a straight line.

(i) \(\angle \) AGE=\(\angle \)GED = \(126^{ 0 }\) 
Alternate Interior Angles
(ii) \(\angle \) GED=\(126^{ 0 }\) 
\(\Rightarrow \angle GEF\) +\( \angle FED\) =\(126^{ 0 }\)
\(\therefore\) EF \(\bot \) CD \(\therefore \) \(\angle \) FED=\(180^{ 0 }\)
\(\Rightarrow \) \(\angle \) GEC=\(126^{ 0 }\) -\(90^{ 0 }\) =\(180^{ 0 }\)
\(\therefore \) EF\(\bot \) CD \(\therefore \) \(\angle \) FED=\(90^{ 0 }\)
\(\angle \)GEF=\(126^{ 0 }\) -\(90^{ 0 }\)=\(36^{ 0 }\)
(iii) \(\angle \) GEC+\(36^{ 0 }\) +\(90^{ 0 }\)=\(180^{ 0 }\)
\(\therefore \)CD is line
\(\Rightarrow \) \(\angle \) GEC+\(36^{ 0 }\) +\(90^{ 0 }\) =\(180^{ 0 }\)
\(\Rightarrow \)  \(\angle \) GEC=\(180^{ 0 }\)-\(126^{ 0 }\)=\(54^{ 0 }\)
Now,  \(\angle \) FGE=\(\angle \) GEC=\(54^{ 0 }\)
Alternate Interior Angles