Construction: Draw ray BL \(\bot \) PQ and ray CM \(\bot \) RS.

BL \(\bot \) PQ,CM \(\bot \) RS and PQ || RS BL||CM
\(\angle \) LBC=\(\angle \) MCB
|Alternate Interior Angles
\(\angle\)ABL=\(\angle\)LBC
Angle of incidence= Angle of reflection
\(\angle \)MCB=\(\angle \)MCD
Angle of incidence= Angle of reflection
From (1),(2) and (3) we get
\(\angle \)ABL =\(\angle \)MCD
Adding (1) and (4) , we get
\(\angle \) LBC+\(\angle \)ABL =\(\angle \)MCB+\(\angle \)MCD
\(\Rightarrow \) \(\angle \)ABC=\(\angle \)BCD
But these form a pair of equal alternate interior angles
So AB || CD.

Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To prove: (i) AD bisects BC (ii) AD bisects \(\angle A\)
Proof: (i) In right \(\triangle ADB\) and right \(\triangle ADC\)
Hyp.AB = Hyp. AC
Side AD = Side AD

\(\triangle ADB\cong \triangle ADC\) | RHS rule
BD = CD | C.P.C.T
AD bisects BC
(ii) \(\triangle ADB\cong \triangle ADC\)
\(\angle BAD=\angle CAD\)
AD bisects \(\angle A\)

Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \(\angle PQR\)
To prove: (i) \(\triangle ABM\cong \triangle PQN\) (ii) \(\triangle ABC\cong \triangle PQR\)
Proof: In \(\triangle ABM\) and \(\triangle PQN\)
AB = PQ
AM = PN
BC = QR
2BM = 2QN | M and Nare the mid-points of BC and QR respectively
BM = QN
In view of (1), (2) and (3)
\(\triangle ABM\cong \triangle PQN\) | SSS rule
(ii) \(\triangle ABM\cong \triangle PQN\)
\(\angle ABM=\angle PQN\) | C.P.C.T
\(\angle ABC=\angle PQR\)
In \(\triangle ABC\) and \(\triangle PQR\)
AB = PQ
BC = QR
\(\angle ABC=\angle PQR\)
\(\triangle ABC=\triangle PQR\) | SAS rule

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.

Given: In figure,
ㄥB < ㄥA and ㄥC < ㄥD.
To Prove: AD < BC
Proof: ㄥB < ㄥA
ㄥA > ㄥB
OB> OA ...(1)
ㄥC < ㄥD
ㄥD > ㄥC
OC > OD  ..(2)
| Side opposite to greater angle is longer
From (1) and 2), we get
OB+OC > OA+OD
⇒ BC> AD
⇒ AD < Be.

Two circles intersect at two points Band C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.
To Prove:\(\angle ACP=\angle QCD\)
Proof:      \(\angle ACP=\angle ABP\)           ...(1)
| Angles in the same segment of a circle are equal
\(\angle QCD=\angle QBD\)          ...(2)
| Angles in the same segment of a circle are equal
\(\angle ABP=\angle QBD\)
| Vertically Opposite Angles
From (1), (2) and (3),
\(\angle ACP=\angle QCD\).             

Given: Circles are drawn with sides AB and AC of a triangle ABC as diameters. They intersect at a point D.
To Prove: D lies on the third side BC of \(\Delta \)ABC
Construction: Join AD

Proof: ∵ Circle drawn on AB as diameter intersects BC in D.
∴  \(\angle ADB=90°\)
| Angle in a semi-circle
But \(\angle ADB+\angle ADC=180°\)
Linear Pair Axiom
∴ \(\angle ADC=90°\)
Hence, the circle described on AC as diameter must pass through D.
Thus, the two circles intersect in D.
Now, \(\angle ADB+\angle ADC=180°\).
∴ Points B, D, C are collinear.
∴ D lies on BC.

125°, 250°, 55°, 60°

45°, 90°, 270°

AB||CF
\(\therefore \angle ABC=\angle BCF\)(Alt Int. \(\angle's\))
\(40=\angle BCF\)
Now, \(\angle ACB+\angle BCF+\angle FCD\)
=180^o(Linear pair)
\(65^0+40^o+\angle FCD=180^o\)
\(\Rightarrow \angle FCD=180^o-105^o\)
\(\Rightarrow \angle FCD=75^o\)
Now, FE||CD
\(\therefore \angle FCD=\angle x\)(Corresponding angles)
\(\angle x=75^o\)

5x-20^o+2x-10^o=180^o (Corresponding interior angles)
\(\Rightarrow\)7x=180^o+30^o=210^o
\(\Rightarrow\)x=30^o
y=180^o-(5x-20^o)
=180^o-(150^o-20^o)
\(\Rightarrow\)y=180^o-130^o=50^o
z=2x-10^o
=60^o-10^o=50^o

\(\angle A=57^o\)
and \(\angle ACD=22^o+35^o=57^o\)
\(\angle A=\angle ACD\)
But these are alternate angles
AB||CD
Again \(\angle FEC+\angle ECD=145^o+35^o=180^o\)
EF||CD
Now AB||CD and EF||CD
AB||EF

Draw, EH||AB

Now, EH||AB and AB||CD,
Therefore, EH||CD
Now, \(\angle BGE+\angle GEH=180^o\) (Co-interior \(\angle S)\)
(AB||EH, Co-interior angles)
\(135^o+\angle GEH=180^o\)
\(\angle GEH=180^o-135^o=45^o...(1)\)
Again, \(\angle DFE+\angle FEH=180^o\)
(CD||EH, Co-interior angles)
\(\Rightarrow 125^o+\angle FEH=180^o\)
\(\Rightarrow \angle FEH=180^o-125^o\)
=55^o
Adding (1) and (2), we get
\(\angle GEH+\angle FEH=45^o+55^o\)
x=100^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
\(\angle 3+\angle 4=(\angle ACB+\angle DCA)+(\angle CAB+\angle DAC)\)
\(=\angle 1+\angle 2\)
\(\therefore \angle 3+\angle 4=\angle 1+\angle 2\)

\(\angle ACD=\angle A+\angle B\)
(Exterior angle is the sum of opp. interior angle)
\(\frac{1}{2}ext.\angle ACD=\frac{\angle A}{2}+\frac{\angle B}{2}\)
\(\angle 2=\angle 1+\frac{1}{2}\angle A...(i)\)
Also, \(\angle 2=\angle 1+\angle E\)
(Exterior angle is the sum of opp. interior angle)
From eqn(1) and (2) we get
\(\therefore \angle1+\frac{\angle A}{2}=\angle 1+\angle E\)
\(\Rightarrow \angle E=\frac{\angle A}{2}\)

We have here \(\angle CAE=\angle EAB(given)...(1)\)
Also \(\angle CAD+\angle DAE=\angle CAE=\angle EAB\) [Use(1)]...(2)
Now \(\angle CAD+\angle C=90^o=\angle DAE+\angle EAB+\angle B\)
\(\angle CAD+\angle C=\angle DAE+\angle CAD+\angle DAE+\angle B\)[use(2)]
\(=2\angle DAE=\angle C-\angle B \)
\(\Rightarrow \angle DAE=\frac{1}{2}(\angle C-\angle B)\)

In \(\triangle\) ABC, as sum of two sides is greater than the 3rd side,
AB + BC>AC ...(1)
In \(\triangle\)ACD, as sum of two sides of a triangle is
greater than the 3rd side,
CD+ DA>AC     ....(2)
Adding (1) and (2), we get
CD + DA + AB.+ BC > 2AC.

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: ABCD is a cyclic quadrilateral whose diagonals intersect at P. \(\angle DBC=70°\) and \(\angle BAC=30°\).
Required: To find \(\angle BCD\) .
Determination: \(\angle BDC=\angle BAC(=30°)\)
| Angles in the same segment of a circle are equal
Now, in \(\Delta BCD\) ,
\(\angle BCD+\angle BDC+\angle DBC=180°\)
| ∵ The sum of the three angles of a \(\Delta \) is 180°
⇒  \(\angle BCD+30°+70°=180°\)
⇒  \(\angle BCD+100°=180°\)
⇒  \(\angle BCD=180°-100°=80°\).

Given: O is the centre of a circle.
To Prove: x + y = z
Proof: \(\angle 3=\angle 4\)
| Angles in the same segment of a circle are equal
\(\angle z=2\angle 3\)
⇒ \(\angle z=\angle 3+\angle 3\)
⇒ \(\angle z=\angle 3+\angle 4\)      .....(1)
Now ​​​​​ \(\angle y=\angle 3+\angle 1\)  .....(2)
   | An exterior angle of a triangle is equal to the sum of its two interior opposite angles
(1) - (2) gives
 \(\angle z-\angle y=\angle 4-\angle 1\)
As \(4=\angle x+\angle 1\Rightarrow \angle 4-\angle 1=\angle x\)
 | An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒ \(\angle 4-\angle 1=\angle x\)         ....(4)
From (3) and (4),
\(\angle z-\angle y=\angle x\)
⇒ \(\angle x+\angle y=\angle z\)
⇒ x+y=z

Given: AB and CD are equal chords intersecting at 90°
To prove: OPQR is a square
Proof: Since P and R are the mid-point of AB and CD respectively
\(\therefore\) \(\angle\) OPB = \(\angle\)ORD = 90°
\(\Rightarrow\)\(\angle\)OPQ = \(\angle\)ORQ = 90° 
Since equal chords on a circle are equidistant from the centre.
\(\therefore\) OP= OR
Thus in t10PQ and t10RQ, we have
OP=OR
\(\angle\)OPQ= \(\angle\)ORQ
and OQ= OQ
\(\therefore\) \(\triangle OPB\cong \triangle ORQ\)
Thus in quadrilateral OPQR,
We have
OP = OR, PQ = RQ
and \(\angle\)OPQ = \(\angle\)ORQ = 90°
Hence OPQR is a square.