\(\angle AGH=\angle GHD\)
\(\Rightarrow \frac{1}{2}\angle AGH=\frac{1}{2}\angle GHD\Rightarrow\angle 1=\angle2\)
\(\Rightarrow\) GM||LH
Similarly, GL||MH
GMHL is a parallelogram.
\(\angle BGH+\angle GHD=180^o\)
\(\Rightarrow\frac{1}{2}\angle BGH+\frac{1}{2}\angle GHD=90^o\)
\(\Rightarrow \angle 3+\angle 2=90^o\)
In \(\triangle GLH, \angle GLH=180^o-(\angle 2+\angle3)\)
=180^o-90^o
\(\Rightarrow \angle GLH=90^o\)
\(\angle GMH=90^o\)
So, \(\angle MGL+\angle GLH=180^o\Rightarrow \angle MGL+90^o\)
=180^o
\(\Rightarrow \angle MGL=90^o\Rightarrow\angle MHL=90^o\)

Given : l||m line t is a transversal intersecting them at P and Q respectively.
To prove : PR||QS
Proof: \(\angle 5=\angle 6\) (Corresponding angles and l||m)
\(\Rightarrow \frac{1}{2}\angle5=\frac{1}{2}\angle6\)
\(\Rightarrow \angle 1=\angle3\)
PR||QS

Given The side BC of \(\triangle \) ABC is produced to D . The bisector of \(\angle A\) meets BC in L.
To prove \(\angle ABC+\angle ACD=2\angle ALC\)
Proof : \(\angle ABC+\angle ACD\)
\(=\angle ABC+(\angle ABC+\angle BAC)\)
|Exterior angle theorem
\(=2\angle ABC+\angle BAC\)
\(=2\angle ABC+2\angle BAL\)
|AL is the bisector of \(\angle A\)
\(=2(\angle ABC+\angle BAL)\)
\(=2\angle ALC\)
|Exterior Angle Theoram

Given ABCD is a quadrilateral
To prove \(\angle A+\angle B+\angle C+\angle D=360^{ 0 }\)
Construction Join AC
Proof In \(\triangle \) ABC

\(\angle 1+\angle B+\angle 3=180^{ 0 }\)
|Angle sum property of a traingleIn 
In \(\triangle \) ADC
\(\angle 2+\angle D+\angle 4=180^{ 0 }\)
|Angle sum property of a traingleAdding (1) and (2) we get
\((\angle 1+\angle 2)+\angle B+(\angle 3+\angle 4)+\angle D=360^{ 0 }\)
\(\Rightarrow \angle A+\angle B+\angle C+\angle D=360^{ 0 }\) 

Given l || m
To prove  \(\angle 1+\angle 2-\angle 3=180^{ 0 }\)
Construction : Through C, draw CF || L || M

\(\therefore l||CF\) | by construction and a transversal BC intersects then
\(\therefore \angle 1+\angle FCB=180^{ 0 }\)
\(\therefore \)The Sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)
\(\Rightarrow \angle 1+\angle FCD=180^{ 0 }\)
BUT \(\angle FCD=\angle 3\)
| Alternate interior angles 
From (1) and (2)
\(\angle 1+\angle 2-\angle 3=180^{ 0 }\) 

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
\(\angle PQT+(\angle QTU+\angle UTS)+\angle RST=360^{ 0 }\) 
\(\Rightarrow \angle PQT+\angle QTS+\angle RST=360^{ 0 }\) 

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

Construction Draw a ray OP opposite to ray OA.
Proof : \(\angle \) AOB+\(\angle \) BOC+\(\angle \) COP+=\(180^{ 0 }\) ...(1)
| \(\because \) A straight angle = \(180^{ 0 }\)
\(\angle \)POD+ \(\angle \)DOE + \(\angle \)EOA =\(180^{ 0 }\) ....(2)
| \(\because \) a straight angle =  \(180^{ 0 }\)
Adding (1) and (2) , we get
\(\angle \)AOB+ \(\angle \)BOC+\(\angle \)COP+\(\angle \)POD+\(\angle \)DOE+\(\angle \)EOA=\(180^{ 0 }+180^{ 0 }=360^{ 0 }\)
\(\Rightarrow \) \(\angle \) AOB+\(\angle \)BOC+\(\angle \)COD+\(\angle \)DOE+\(\angle \)EOA=\(360^{ 0 }\).