(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.

DE||PR and DE=\(\frac{1}{2}\)PR (mid-point theorem)
EF||PQ and EF=\(\frac{1}{2}\)PQ
DF||QR and DF=\(\frac{1}{2}\)QR
As PQ= QR = PR
(PQR is an equilateral triangle)
⇒ DE=EF=DF
⇒ ΔDEF is an equilateral triangle.

Join diagonal AC
In ΔADC,
S is the mid-point of AD.
R is the mid-point of DC.

SR || AC
and SR =\(\frac{1}{2}\)AC (By mid-point theorem)...(1)
Similarly in ΔBAC,
PQ || AC
and PQ=\(\frac{1}{2}\) AC...(2)
From (1) and (2),
SR II PQ
and SR = PQ
ஃ PQRS is a parallelogram.

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)

Let PQRS be a rectangle
In ΔSQP and ΔRQP, 
SP = RQ, (Opp. sides of rectangle are equal)
ㄥSQP=ㄥRQP=90°
(Each angle of rectangle is right angle)
PQ=QP (common)
ΔSQP≡ΔRQP (By SAS Congruence Rule)
QS = PR (by c.p.c.t.)
Hence Proved.

55°

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 parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively.
To Prove:
\((i)\Delta APB\cong \Delta CQD\)
\(\\ (ii)AP=CQ\)
Proof: (i) In \(\Delta APB\ \) and \( \Delta CQD\)
AB = CD                I Opp. sides of || gm ABCD
\(\angle ABP=\angle CDQ\)           |AB || DC and transversal BD intersects them
 \(\angle ABP=\angle CDQ\)           |Each=\(90°\)
(ii)  \(\Delta APB\cong \Delta CQD\)                I Proved above in (i)
 \(\therefore \)  AP = CQ.                I C.P.C.T.

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: 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.