Let \(\angle DAF=\angle 1\)
\(\angle CFA=\angle 2,\)
\(\angle BFE=\angle 3\)
\(\angle BEF=\angle 4\)
Since, AE||DC
\(\angle D+\angle 1=180^o\)
(Angles on the same side of transversal)
\(\angle 1=180^o-110^o=70^o\)
\(\angle 4=\angle 1=70^o\) (Alternate angle)
Again, \(97^o+\angle2=180^o\) (Angle on the same side of transversal)
\(\angle 2=180^o-97^o=83^o\)
\(\angle 3=\angle2=83^o\)
(Vertically opp.angles)

In \(\triangle BEF,\)
\(\angle 3+\angle 4+\angle EBF=180^o\)
(Angle sum property)
\(\Rightarrow 83^o+70^0+\angle EBF=180^o\)
\(\Rightarrow \angle EBF=180^o-153^o\)
\(\Rightarrow \angle EBF=27^o\)

AB||DC
\(\angle CDB=\angle ABD\)
=x=35^o[alternate angles]
x+y+80^o=180^o
\(\angle ADB=y=180^o-35^o-80^o\)
=65[angle sum property]
\(\angle DCB=z=180^o-[35^o+35^o]\)
=110^o

Given PQ || RS and T is any point
To prove \(\angle PQT+\angle QTS+\angle RST=360^{ 0 }\)
Construction Through T, draw TU || PQ || RS

PQ || UT  |By construction and a transversal QT intersects then
\(\therefore \angle PQT+\angle QTU=180^{ 0 }\)
The Sum of consecutive interior angles on the same sides of a transversal is \(180^{ 0 }\)
UT || RS
| By construction and a transversal TS intersect them
\(\therefore \angle UTS+\angle RST=180^{ 0 }\)
The Sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)
Adding (1) and (2) ,we get
\(\angle PQT+(\angle QTU+\angle UTS)+\angle RST=360^{ 0 }\) 
\(\Rightarrow \angle PQT+\angle QTS+\angle RST=360^{ 0 }\) 

Through O draw OE || AB || CD 
Now y= \(\angle FOG\)
=\(\angle \)FOE + \(\angle GOE\)
=\(\angle \)CFO+\(\angle \) AGO
=\(\angle \)FOE ==\(\angle \) CFO (Alternate Interior angles)
=\(\angle \)GOE=\(\angle \)AGO (Alternate Interior Angles)
 =\(45^{ 0 }\)+\(40^{ 0 }\)=\(85^{ 0 }\)

Given l,m,n are three lines such that l || m and  n \(\bot \)m

\(\therefore \)l || m and n is a transversal
\(\therefore \) \(\angle \)1=\(\angle \)2 |Corresponding angles
But \(\angle \)1=\(90^{ 0 }\)
\(\therefore \) \(\angle \) 2= \(90^{ 0 }\)
\(\Rightarrow \) n \(\bot \) m

Let the two lines m and n each be perpendicular to the same line L

Then,
\(\angle \)1=\(90^{ 0 }\)
\(\angle \)2=\(90^{ 0 }\)
\(\angle \)1=\(\angle \)2
But these angles form a pair of equal corresponding angles
\(\therefore \) m||n

\(\therefore \) ROS is a line 
\(\therefore \) 4b+\(75^{ 0 }\)+b=\(180^{ 0 }\)
\(\Rightarrow 5b=180^{ 0 }-75^{ 0 }=105^{ 0 }\)
\(\Rightarrow \) b=\(\frac { 105^{ 0 } }{ 5 } =21^{ 0 }\)
2c=\(75^{ 0 }\)+b
|Vertically opposite angles
\(\Rightarrow 2c=75^{ 0 }+21^{ 0 }\)
\(\Rightarrow 2c=96^{ 0 }\)
\(\Rightarrow \)\(c=\frac { 96^{ 0 } }{ 2 } =48^{ 0 }\)
a=4b
|Vertically opposite angkes 
\(\Rightarrow a=4X21^{ 0 }=84^{ 0 }\)
Thus \(a=84^{ 0 },b=21^{ 0 },c=48^{ 0 }\) 

Let AB and CD two lines intersecting at O

This leads to two pairs of vertically  opposite angles,namely
(i)  \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
We are to prove that
(i)   \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
\(\therefore \) Ray OA stands on line CD 
Therefore (i)   \(\angle \)AOC and \(\angle \)AOD=\(180^{ 0 }\)
|Linear Pair Axiom
From (1) and (2) 
\(\angle \)AOC and \(\angle \)AOD= \(\angle \)AOD and \(\angle \)BOD \(\angle \)AOC and \(\angle \)BOD
\(\Rightarrow \) Similarly we can prove that 
\(\angle \)AOD and \(\angle \)BOC

\(\therefore \) PO \(\bot \) AB
\(\therefore \angle AOP=90^{ 0 }\)
x:y:z=1:3:5=9
\(\therefore \angle x=\frac { 1 }{ 9 } x90^{ 0 }=10^{ 0 }\)
\(y=\frac { 3 }{ 9 } x90^{ 0 }=30^{ 0 }\)
\(y=\frac { 5 }{ 9 } x90^{ 0 }=50^{ 0 }\)