Given Two lines AB and CD.
Intersect at a point O.

To prove
\((i)\angle AOC=\angle BOD\)
\((ii)\angle AOD=\angle BOC\)
Proof Ray OA stand on line CD
\(\therefore \angle AOC+\angle AOD=180^o\) ...(i) [Linear pair]
Again ray OD stand on line AB
\(\therefore \angle AOD+\angle BOD=180^o\) ..(ii) [Linear pair]
from eqn (i) and (ii)
\(\angle AOC+\angle AOD=\angle AOD+\angle BOD\) [Each equal to 180^o]
\(\therefore \angle AOC=\angle BOD\)
Similarly \(\angle AOD=\angle BOC\)

Given\(\angle ABC=\angle ACB\)
\(\Rightarrow \angle1+\angle4=\angle2+\angle3\)
\(\Rightarrow \angle1+\angle4-\angle4=\angle2+\angle3-\angle3\)
(As, \(\angle3=\angle4\))
\(\Rightarrow \angle1=\angle2\)

The equation is y=9x-7
A(1,2); 2=9(1)-7
2=2; true
B(-1,-16); -16=9(-1)-7=-9-7
=-16; True
C(0,-7); -7=9(0)-7
=0-7=-7; True

Given: Sides AB and AC of ΔABC are extended to points P and Q respectively.Also, ㄥPBC < ㄥQCB.
To Prove: AC > AB.
Proof: ㄥPBC < ㄥQCB
- ㄥPBC > -ㄥQCB
180⁰ - ㄥ PBC > 180⁰- ㄥQCB
ㄥABC > ㄥACB
AC > AB.

Parallelogram; yes

7

\(4\sqrt { 5 } +5\sqrt { 2 } \)

See a blackboard

ㄥA > ㄥB > ㄥC
BC > AC

0=8\((90^{ 0 }-\theta )\)

Value=\(6(1)^2+7(1)-3\)
\(​​​​​​​=6+7-3=10\)

To prove: Sum of all the angles of \(\triangle ABC\) is 180^o
Construction : Draw a line l parallel to BC.
Proof: Since \(l||BC\) , we have \(\angle 2=\angle y\) (Alternate angles are equal)...(i)
Similarly, \(l||BC\) \(\angle 1=\angle z\) (Alternate angles are equal)...(ii)
Also, sum of angles at a point A on line l is 180^o
\(\therefore \angle 2+\angle x+\angle 1=180^o\) (linear pair)
i.e., \(\angle y+\angle x+\angle z=180^o\) (from (i) and(ii)).
\(\therefore\ \angle x+\angle y+\angle z=180^o\)
Sum of all angles of a \(\triangle \) is 180^o
5x+6x+7x=180^o
x=10^o
Angles are 50^o, 60^o and 70^o respectively.
Hence proved.

Given: ABC is an isosceles triangle in which AB = AC. AD bisects L PAC and CD IIAB.
To Prove:
(i)  \(\angle \)DAC =\(\angle \)BCA
(ii) ABCD is a parallelogram.
Proof:
(i) In \(\Delta \) ABC,
\(\because\) AB = AC
\(\therefore\) \(\angle \)B =\(\angle \)C               .......(1)      I Angles opposite to equal sides of a triangle are equal
Also, Ext.   \(\angle \)PAC =\(\angle \)B +\(\angle \)C     I An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒    \(\angle \)PAC = \(\angle \)C+\(\angle \)C   | From (1)
⇒ 2\(\angle \)CAD = 2\(\angle \)C   | \(\because\) AD bisects \(\angle \)PAC 
⇒ \(\angle \)CAD =\(\angle \)C
⇒  \(\angle \) DAC =\(\angle \)BCA
(ii) But these angles form a pair of equal alternate interior angles
\(\therefore\)  AD II BC
Also, CD II AB         I Given
\(\therefore\)   ABCD is a parallelogram   IA quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel.

Given: A \(\triangle ABC\) in which the bisector of the vertical angle \(\angle BAC\) bisects the base BC, i.e., BD = CD
To prove: \(\triangle ABC\) is isosceles
Construction: Produce AD to E such thatAD = DE. Join EC.
Proof: In \(\triangle ADB\) and \(\triangle EDC\)
BD = CD
AD = ED
\(\angle ADB=\angle EDC\) | Vertically opposite angles
\(\triangle ADB=\triangle EDC\) | SAS congruence rule
AB = EC ...... (1) | C.P.C.T
\(\angle BAD=\angle CED\) | C.P.C.T
But \(\angle BAD=\angle CAD\)
\(\angle CAD=\angle CED\)
AC = AE ..... (2) | Sides opposite to equal angles of a triangle are equal
From (1) and (2)
AB = AC
\(\triangle ABC\) is isosceles.

x = 3 represents a line parallel to y-axis at a distance of 3 units to the right of the origin.
4x= 3y
\(\Rightarrow\ \ \ y={4x\over 3}\)
Table of solution

We plot the points (0,0) and (3, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x = 3y.

Area of the triangle GAB formed by the given two lines and the x-axis \(={3\times 4\over2}=6\) square units

x+2y=8
2y=8-x
\(y={8-x\over 2}\)
This expresses y in terms of x
This line will intersect x-axis at the point for while y=0.So, put y=0 in (1), we get
x+2(0)=8
x=8
Hence, line(1) intersects x-axis at the point (8, 0).
This line will cut y-axis at the point for which x=0.So, put x=0 in (1), we get
0+2y=8
2y=8
\(y={8\over2}=4\)
Hence, line(1) cuts y-axis at the point (0, 4).

\({ \left( x-\frac { 2 }{ 3 } y \right) }^{ 3 }-{ \left( x+\frac { 2 }{ 3 } y \right) }^{ 3 }\)
\(=\left[ { x }^{ 3 }-{ \left( \frac { 2 }{ 3 } y \right) }^{ 3 }-3(x){ \left( \frac { 2 }{ 3 } y \right) }\left\{ (x)-{ \left( \frac { 2 }{ 3 } \right) } \right\} \right] \)
\(=\left[ { x }^{ 3 }-{ \left( \frac { 2 }{ 3 } y \right) }^{ 3 }-3(x){ \left( \frac { 2 }{ 3 } y \right) }\left\{ (x)+{ \left( \frac { 2 }{ 3 } \right) } \right\} \right] \)
\(=\left[ { x }^{ 3 }-\frac { 8 }{ 27 } { y }^{ 3 }-{ 2x }^{ 2 }y+\frac { 4 }{ 3 } { xy }^{ 2 } \right] -\left[ { x }^{ 3 }+\frac { 8 }{ 27 } { y }^{ 3 }+{ 2x }^{ 2 }y+\frac { 4 }{ 3 } { xy }^{ 2 } \right] \)
\(=-\frac { 16 }{ 27 } { y }^{ 3 }-4{ x }^{ 2 }y \) 

\(\angle QPR=75^o\) (Vertically opposite angles)
Again, \(\angle PQR+\angle QPR=105^o\) (Exterior angle)
\(\Rightarrow \angle PQR+75^o=105^o\)
\(\Rightarrow PQR=30^o\)

OA = OB (O is the mid-point of AB)
\(\angle\)AOC = \(\angle\)BOD (Vertically opposite angles)
OC = OD (O is the mid-point of CD)
\(\triangle AOC\cong \triangle BOD\)
\(\Rightarrow\) AC = BD. (By c.p.c.t) Proved.

AB = CD(Given)
\(\Rightarrow\)AB + BC = BC + CD
\(\Rightarrow\)AC = BD
Euclid's axiom used: If equals are added to equals, the wholes are equal.

Euclid's axioms
(i) Things which are equal to the same thing are equal to one another.
(ii) If equals are added to equals, the wholes are equal.

x=8

(2,3), (2,0), (0,3)