Given, PQRS is a parallelogram.
ㄥP+ㄥQ=180°
\(\frac{1}{2}\)ㄥP+\(\frac{1}{2}\)ㄥQ=\(\frac{180}{2}\)=90°
ㄥPTQ=180°-{\(\frac{1}{2}\)(ㄥP+ㄥQ)}
=180°-90°
ㄥPTQ=90°

In Δs RSM and PQL, 

ㄥRSM=ㄥPQL
ㄥM=ㄥL=90°
SR=PQ
By AAS, ΔRSM≡ΔPQL
(ii) PL=RM(c.p.c.t)

Given: ABCD is a trapezium in which AB II DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F.
To Prove: F is the mid-point of Be.
Proof: Let DB intersect EF at G.
In \(\Delta \)DAB,
\(\because\) E is the mid-point of DA and EG II AB
\(\therefore\) G is the mid-point of DB     | By converse of mid-point theorem
Again, in  \(\Delta \)BDC, 
 \(\because\)  G is the mid-point of BD and GF II AB II DC
 \(\therefore\)  F is the mid-point of BC.   | By converse of mid-point theorem

Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.
To Prove: Quadrilateral ABCD is a square.
Proof: In \(\Delta\)OAD and \(\Delta\)OCB,
OA = OC             I Given
OD = OB       I Given
\(\angle AOD=\angle COB\)    |Vertically Opposite Angles
 \(\therefore \Delta OAD\cong \Delta OCB\)    I SAS Congruence Rule

\(\therefore \)   AD = CB           I C.P.C.T.
 \(\angle ODA=\angle OBC\)  |C.P.C.T
\(\angle BDA=\angle DBC\)
.\(\therefore \)   AD II BC
Now, \(\because\) AD = CB and AD II CB
\(\therefore \)  Quadrilateral ABCD is a II gm.      I A quadrilateral is a parallelogram if a pair of opposite sides are parallel and  equal.
In \(\therefore \) \(\Delta\) AOB and \(\Delta\)AOD,
AO = AO                          I Common
OB = OD          I Given
 \(\angle \)AOB = \(\angle \)AOD               |Each=\(90°\)
\(\therefore \)  \(\Delta\)AOB \(\cong \)  \(\Delta\) AOD         I SAS Congruence Rule
AB = AD                 I C.P.C.T.
Now, \(\because\) ABCD is a parallelogram and
AB=AD
 \(\because\) ABCD is a rhombus.
Again, in \(\Delta\) ABC and \(\Delta\) BAD,
AC = BD            I Given
BC = AD            I \(\because\)  ABCD is a rhombus
AB =  BA                     I Common
\(\therefore \) \(\Delta\) ABC \(\cong \) \(\Delta\) BAD       I SSS Congruence Rule
\(\therefore \)    \(\angle \) ABC = \(\angle\) BAD    I C.P.C.T.
\(\because\)    AD II BC                 I Opp. sides of IIgm ABCD and transversal AB intersects them.
\(\therefore \) \(\angle ABC+\angle BAD=180°\)      I Sum of consecutive interior angles on the  same side of a transversal is 180°
ABC = \(\angle\)BAD = \(90°\)
Similarly,\(\angle\)BCD =\(\angle\)ADC =\(90°\)
 \(\therefore \)  ABCD is a square.

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA = 90°
⇒ ㄥLON = 90°
Similarly, ㄥOLM =ㄥLMN =ㄥMNO = 90°

ஃ A quadrilateral with all angles 90° is a rectangle. Also opposite angles are equal. It is rectangle.

Given: ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively.
To Prove: AX IICY.
Proof: \(\because\) ABCD is a parallelogram.
\(\therefore\) \(\angle \)A = \(\angle \)C  I Opposite \(\angle \)s of a parallelogram are equal

⇒  \(1\over 2\)\(\angle \)A=\(1\over 2\)\(\angle \)C     |\(\because \)Halves of equals are equal
⇒    \(\angle \)1 = \(\angle \)2    ...(1)      I    \(\because\)   AX is the bisector of \(\angle \)A and CY is the bisector of \(\angle \)C Now, AB II DC and CY intersects them
 \(\therefore\)  \(\angle \)2 =\(\angle \)3  ....(2)   I Alternate interior \(\angle \) s 
From (1) and (2), we get
\(\angle \)1 = \(\angle \)3
But these form a pair of equal corresponding angles
\(\therefore\)  AX II CY.

Given: In a quadrilateral ABCD, the line segments bisecting \(\angle \)C and \(\angle \)D meet at E.
  To Prove:   \(​​\angle \)A+\(​​\angle \)B = 2 \(\angle \)CED
I Angle sum property of a quadrilateral
In \(\Delta \) CED,
\(\angle \)CED +\(\angle \)EDC+\(\angle \)ECD = \(180°\)   | Angle sum property of a triangle
⇒ \(\angle \)CED+\(1\over2\)\(\angle \)D+\(1\over2\)\(\angle \)C=\(180°\)
 ⇒ 2\(\angle \)CED +\(\angle \) D+\(\angle \)C =\(360°\)    ...(2)
From (1) .and (2),
2\(\angle \)CED+\(\angle \)D+C=A+B+C+D
 ⇒ 2 \(\angle \)CED =\(\angle \)A+\(\angle \)B
 ⇒ \(\angle \)A+\(\angle \)B=2\(\angle \)CED

AM丄DC, AN丄BC
In quadrilateral AMCN,
ㄥA+ㄥM+ㄥC+ㄥN=360°
ㄥA+ㄥC=180°
⇒ 50°+ㄥC=180° ⇒ ㄥC=130°

In parallelogram, ㄥA=ㄥC=130°
ㄥB=ㄥD=180°-130°
=50°

∠APR=ㄥDRP
or ㄥ1=ㄥ2
But these are alternate interior angles
SP II RQ, SR II PQ
PQRS is a parallelogram
∠APR+ㄥBPR=180°,(linear pair)
⇒ \(\frac{1}{2}\) ∠APR+\(\frac{1}{2}\)ㄥBPR=\(\frac{1}{2}\)x180°
⇒ ∠1+ㄥ3=90°
⇒ ∠SPQ=90°

PQRS is a rectangle

Let the measure of the angles be x, 2.x, 3x and 4x then,
x + 2x + 3x + 4x = 360°
⇒ x = 36°
ஃ Angles of quadrilateral are 36°, 72°, 108°,144°

72°, 108°, 72°, 108°

\(36°\),\(144°\),\(36°\), \(144°\)

\(\angle \)DAB = \(\angle \)DCB = 60°

ㄥR=80° (Given)
SR II PQ and RQ is a transversal
ஃ ㄥR+ㄥQ=180°
ㄥQ=180°-80°
= 100°
Similarly, ㄥQ+ㄥP=180°
⇒ ㄥP=180°-100°=80°
and ㄥS+ㄥR=180°
⇒ ㄥS=180°-80°=100°
Hence, ㄥP=80°, ㄥQ=100°, ㄥR=80°, ㄥS=100°

(ii) Property of co-interior angles when a pair of straight lines intersected by another straight line (Geometry)
(iii) Diligence i.e., dedication, determination, and hard work.

(i) As EB II DL and ED II BL.
Therefore EBLD is a parallelogram.
ஃ BL = ED
=\(\frac{1}{2}\)BC=CL ...(i)
Now in triangles DCL and FBL, we have
CL=BL from (i)
ㄥDLC= ㄥFLB 
ㄥCDL=ㄥBFL
ΔCDL≡ΔBFL
CD= BF
and DL= FL
Now, BF = DC = AB
⇒ 2AB = 2DC
⇒ AF = 2DC
(ii) ∵ DL=FL
⇒ DF = 2DL