We have a square ABCD such that its diagonals AC and BD intersect at O.
(i) To prove that the diagonals are equal, i.e. AC = BD
In ΔABC and ΔBAD, we have
AB =BA
[Common] 
BC = AD [Opposite sides oj the square ABCD]
∠ABC = ∠BAD [All angles of a square are equal to 90^ο]
∴ ABC ≅ ΔBAD [SAS criteria]
⇒ Their corresponding parts are equal.
⇒  AC = BD ...(1)
(ii) To prove that '0' is the mid-point of AC and BD.
∵ AD II BC and AC is a transversal. [∵ Opposite sides of a square are parallel]
∴ ∠1 = ∠3 [Interior alternate angles]
Similarly, ∠2 = ∠4 [Interior alternate angles]
Now, in ΔOAD and ΔOCB, we have
AD = CB [Opposite sides oj the square ABCD]
∠1 = ∠3 [Proved]
∠2 = ∠4 [Proved]
∴ ΔOAD == ΔOCB [ASA criteria]
∴ Their corresponding parts are equal.
⇒ OA = OC and OD = OB
⇒ O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O. ...(2)
(iii) To prove that AC ⏊ BD.
In ΔOBA and ΔODA, we have
OB = OD [Proved]
BA = DA [Opposite sides of the square]
OA = OA [Common]
∴ ΔOBA ≅ ΔODA [SSS criteria]
⇒ Their corresponding parts are equal.
⇒ ∠AOB = ∠AOD
But ∠AOB and ∠AOD form a linear pair.
∴ ∠AOB + ∠AOD = 180^ο
⇒ ∠AOB = ∠AOD = 90^ο
⇒ AC ⏊ BD ... (3)
From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.

(i) Since, ABCD is a parallelogram
AB = CD and AB II CD
⇒ \(\frac{1}{2}\)AB=\(\frac{1}{2}\)CD
i.e., AP = CQ and AP II CQ
⇒ APCQ is a parallelogram
(ii) Again
 \(\frac{1}{2}\)AB=\(\frac{1}{2}\)CD
i.e., PB = DQ and PB II DQ
⇒ DPBQ is a parallelogram
(iii) QS II PR and SP II QR
⇒ PSQR is a parallelogram.

In ΔABC, P and Q are midpoints of AB and BC respectively
ஃ PQ=\(\frac{1}{2}\)AC and PQ IIAC ...(i)
Similarly, RS=\(\frac{1}{2}\)AC and RS IIAC ...(ii)
ஃ PQRS is a parallelogram
Also AD= BC ⇒ AS = BQ
In ΔAEH and ΔBEF,
AP = BP,AS = BQ and ㄥA=ㄥB=90
ஃ ΔAPS≡ΔBPQ
⇒ PS = PQ ...(iii)
From (i) and (iii), PQRS is a parallelogram with
PQ=PS
i.e., PQRS is a rhombus.
Hence proved.

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°

A parallelogram PQRS in which the bisectors of LP and LQ meet SR at O.
To Prove: ㄥPOQ=90°
Now, since PQRS is a parallelogram. Therefore,
PSIIQR
Now, PS II QR and transversal PQ intersects them.
ㄥP+ㄥQ=180°
(∵ Sum of consecutive interior angles is 180°)
\(\frac{1}{2}\)ㄥP+\(\frac{1}{2}\)ㄥQ=90°
⇒ ㄥ1+ㄥ2=90° ( OP is bisector of ㄥP...(i) and OQ is bisector of ㄥQ.
ㄥ1=\(\frac{1}{2}\) ㄥP and ㄥ2=\(\frac{1}{2}\) ㄥQ)
Now, in ΔPOQ
ㄥ1+ㄥPOQ+ㄥ2=180°
⇒ 90°+ㄥPOQ=180°
⇒ ㄥPOQ=90°

In Δs APD and CQB
AD = BC (Opp. sides of a parallelogram)
PD = BQ (Given)
∠ADP=∠QBC
⇒ ΔAPD≡ΔCQB
⇒ AP = CQ (c.p.c.t)
In Δs AQB and CPD
AB = DC, BQ = DP
and ∠AQB=∠PDC
ஃ ΔAQB≡ΔCPB ⇒ AQ=CP
(iii) In quad. APCQ, 
AP = CQ and AQ = CP
⇒ APCQ is a parallelogram.

In Δs RSM and PQL, 

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

In a parallelogram ABCD, AO = OC
[\(\because\)  diagonals of a parallelogram bisect each other]
AC = AO + OC
\(\Rightarrow\)  AC = 2OC
\(\therefore CQ=\frac { 1 }{ 4 } AC=\frac { 1 }{ 4 } \times \left( 2OC \right) =\frac { 1 }{ 2 } OC\)
Thus, Q is the mid-point of OC.
Now, in \(\triangle CDO\), P and Q are the mid-points of CD and CO, respectively.
\(\therefore PQ\parallel DO\) and therefore, \(QR\parallel OB\)
[by mid-point theorem]
\(\left[ \because PQ\parallel DO\Rightarrow PQR\parallel DOB \right] \)
Now, in \(\triangle COB\), Q is the mid-point of CO and \(QR\parallel OB\)
[by the converse of mid-point]
So, R must be the mid-point of BC.

Given, parallelogram ABCD, in which \(\angle ADC\) and \(\angle ABC\) are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.
Now, BEDF is a quadrilateral, in which
\(\angle BED=\angle BFD\) = 90°
\(\therefore \angle FBE={ 360 }^{ ° }-\left( \angle FDE+\angle BED+\angle BFD \right) \)
[angle sum property of a quadrilateral]
= 360° - (60° + 90° + 90°)
=360° - 240° = 120°
Since, ABCD is a parallelogram.
\(\therefore \angle ADC\) = 120° [\(\because \angle D=\angle B\)]
Now, \(\angle A+\angle B\) = 180°
[co-interior angles of a parallelogram]
\(\therefore \angle A={ 180 }^{ ° }-\angle B={ 180 }^{ ° }-{ 120 }^{ ° }\)
[\(\because\angle FBE=\angle B\)]
\(\Rightarrow \angle A\) = 60°
Also, \(\angle C=\angle A\) = 60°
[\(\because\)  opposite angles of a parallelogram are equal]
Hence, angles of the parallelogram are 60°, 120°, 60°, and 120° respectively.

Lines l and m are parallel, i.e. PS\(\parallel \)QR and transversal t intersects PS and QR at points A and C, respectively. The bisectors of \(\angle\)PAC and \(\angle\)ACQ intersect each other at B and bisectors of \(\angle\)ACR and \(\angle\)SAC intersect each other at D. We have to show that, quadrilateral ABCD is a rectangle.
Since PS\(\parallel \)CR and t is transversal.
\(\therefore\)  \(\angle\)PAC =\(\angle\)ACR  [alternate angles]
\(\Rightarrow \frac { 1 }{ 2 } \angle PAC=\frac { 1 }{ 2 } \angle ACR\)
\(\Rightarrow \angle BAC=\angle ACD\)
These form a pair of alternate angles for lines AB and CD with AC as transversal.
So,  AB\(\parallel \)DC
Similarly, BC\(\parallel \) AD
Therefore, quadrilateral ABCD is a parallelogram.
Now, PS is a straight line.
So,  \(\angle\)PAC + \(\angle\)SAC = 180\(°\)
[linear pair axiom]
\(\Rightarrow \frac { 1 }{ 2 } \angle PAC+\frac { 1 }{ 2 } \angle SAC=\frac { 1 }{ 2 } \times 180°\)
\(\left[ multiply\ by\frac { 1 }{ 2 } on\ both\ sides \right] \)
\(\Rightarrow \angle BAC+\angle DAC=90°\)
\(\Rightarrow \angle BAD=90°\)
Thus, one angle of parallelogram ABCD is a right angle.
Hence, ABCD is a rectangle.                          
Hence Proved.

Given: ABCD is a quadrilateral. P, Q, R, and S are the mid-points of the sides DC, CB, BA, and  AD respectively.
To Prove: PR and QS bisect each other.

Construction: Join PQ, QR, RS, SP, AC and BD
Proof: In \(\Delta\)ABC,
\(\because\) R and Q are the mid-points of AB and BC respectively.
\(\therefore\) RQ II AC and RQ = \(1\over2\)AC.
Similarly, we can show that
PS II AC and PS = \(1\over2\) AC
\(\therefore\) RQ II PS and RQ = PS.
Thus a pair of opposite sides of a quadrilateral PQRS are parallel and equal.
\(\therefore\)  PQRS is a parallelogram.
Since the diagonals of a parallelogram bisect each other.
\(\therefore\) PR and QS bisect each other.

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: Diagonal AC of a parallelogram ABCD bisects \(\angle A\)
To Prove:
(i) It bisects \(\angle C\) also
(ii) ABCD is a rhombus.
Proof: (i) In \(\Delta\)ADC and \(\Delta\)CBA,
AD = CB                 I Opp. sides of IIgm ABCD
CA = AC               I Common
DC = BA               I Opp. sides of II gm ABCD
\(\therefore \)  \(\Delta\)ADC \(\cong \) \(\Delta\)CBA               I SSS Congruence Rule
\(\angle ACD=\angle CAB\)     |C.P.C.T
 \(\angle DAC=\angle BCA\)     |C.P.C.T
but  \(\angle CAB=\angle DAC\)     |Given
 \(\therefore \ \angle ACD=\angle BCA\)
 \(\therefore \) AC bisects \(\angle C\) also
(ii) From above,
\(\angle ACD=\angle CAD\)  
AD = CD            I Sides opposite to equal angles  of a triangle are equal
\(\therefore \) AB = BC = CD = DA         I \(\therefore \) ABCD is a II gm
\(\therefore \)  ABCD is a rhombus.

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.

Given: In parallelogram ABCD, AC = BD   To Prove: ||gm ABCD is a rectangle.

Proof: In \(\Delta\) ACB and \(\Delta\)BDA,
AC = BD                                 I Given
AB = BA                                 I Common
BC = AD                                 I Opposite sides of II gm ABCD
\(\therefore\) \(\Delta\)ACB =\(\Delta\)BDA               | SSS Congruence Rule
\(\therefore\)   \(\angle ABC=\angle BAD\quad \quad \quad ......(1)\quad C.P.C.T\) 
AD || BC                    | Opp. sides of II gm ABCD
and transversal AB intersects them.
 \(\angle BAD+\angle ABC=180°\quad \quad \quad ....(2)\)
|Sum of consecutive interior angles on the same side of a transversal is 180
From (1) and (2),
 \(\angle BAD=\angle ABC=90°\)
\(\therefore\) ​​​​​|| gm ABCD is a rectangle.

90°

∠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

62

Given ABCD is a parallelogram and AP = CQ
To prove AC and PQ bisect each other.
Proof In \(\triangle AMP\) and \(\triangle CMQ\),
\(\angle MAP=\angle MCQ\) [alternate interior angles]
AP = CQ [given]
\(\angle APM=\angle CQM\) [alternate interior angles]
\(\therefore \triangle AMP\cong \triangle CMQ\)  [by ASA congruence rule]
Then, AM = CM   [by CPCT]
and PM = QM [by CPCT]
So, AC and PQ bisect each other.

10.5 cm

5 cm

72°, 108°, 72°, 108°