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P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

Given: In a parallelogram ABCD, P is the mid-point of DC
To Prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram
 BC = AD and BC\parallel AD
Also, DC = AB and DC\parallel AB
P is the mid-point of the DC
DP=PC=\frac{1}{2}DC
Now QC\parallel AP and PC\parallel AQ
So APCQ is a parallelogram
AQ=PC=\frac{1}{2}DC

\frac{1}{2}AB=BQ         …..(1) {\because DC = AB}
In \triangle AQR  &  \triangle BQC AQ = BQ {from equation 1}
\angle AQR=\angle BQR {vertically opposite angles}
\angle ARQ=\angle BCQ {alternate angles of transversal}
\triangle AQR\cong \triangle BQC {AAS congruence}
AR = BC {by CPCT}
BC = DA
AR = DA
Also, CQ = QR {by CPCT}
Hence Proved.

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Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Given: Let ABCD be a trapezium in which AB\parallel DC  and let M and N be the mid-points of diagonals AC and BD.

To Prove: MN\parallel AB\parallel CD
Proof: Join CN and produce it to meet AB at E
In \triangle CDN and \triangle EBN we have
DN = BN {N is the mid-point of BD}
\angle DCN=\angle BEN {alternate interior angle}
\angle CDN=\angle EBN {alternate interior angles}
\triangle CDN\cong \triangle EBN 
DC = EB and CN = NE {by CPCT}
Thus \triangle CAE, the points M and N are the mid-points of AC and CE, respectively.

MN\parallel AE   {By mid-point theorem}
MN\parallel AB\parallel CD
Hence Proved

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D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

Given: In \triangle ABC, D, E and F are respectively the mid-points of the sides AB, BC and CA.
To prove: \triangle ABC is divided into four congruent triangles.

Proof: Using the given conditions we have
AD=BD=\frac{1}{2}AB,BE=EC=\frac{1}{2}BC,AF=CF=\frac{1}{2}AC
Using mid-point theorem
EF\parallel AB  and   EF=\frac{1}{2}AB=AD=BD
ED\parallel AC  and  ED=\frac{1}{2}AC=AF=CF
DF\parallel BC  and   DF=\frac{1}{2}BC=BE=EC
In \triangle ADF and \triangle EFD
 AD = EF
 AF = DE
 DF = FD {common side}
\triangle ADF\cong \triangle EFD  {by SSS congruence}
Similarly, we can prove that,
\triangle DEF\cong \triangle EDB
\triangle DEF\cong \triangle CFE
So, \triangle ABC is divided into four congruent triangles
Hence proved

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ABCD is a rectangle in which diagonal BD bisects \angle B. Show that ABCD is a square.

Given: In a rectangle ABCD, diagonal BD bisects B
To Prove: ABCD is a square
Construction: Join AC

Proof:
Given that ABCD is a rectangle. So all angles are equal to 90^{\circ}
Now, BD bisects \angle B
\angle DBA=\angle CBD
Also,
\angle DBA+\angle CBD=90^{\circ}  
So,
2\angle DBA=90^{\circ}


\angle DBA=45^{\circ}
In \triangle ABD,
\angle ABD+\angle BDA+\angle DAB=180^{\circ}  
(Angle sum property)
45^{\circ}+\angle BDA+90^{\circ}=180^{\circ}
\angle BDA=45^{\circ}
In \triangle ABD,
AD = AB (sides opposite to equal angles in a triangle are equal)
Similarly, we can prove that BC = CD
So, AB = BC = CD = DA
So ABCD is a square.
Hence proved

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P and Q are points on opposite sides of AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To Prove: PQ is bisected at O.


Proof: In \triangle ODP and \triangle OBQ
\angle BOQ=\angle POD {Vertically opposite angles}
\angle OBQ=\angle ODP {interior angles}
OB = OD {given}
\triangle ODP\cong \triangle OBQ {by ASA congruence}
OP = OQ {by CPCT rule}
So, PQ is bisected at O.

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Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Given: Let ABCD be a parallelogram and AP, BR, CQ, and DS are the bisectors of
\angle A, \angle B, \angle Cand \angle D respectively.
To Prove: Quadrilateral PQRS is a rectangle.

Proof: Since ABCD is a parallelogram
Then DC\parallel AB and DA is transversal.
\angle A+\angle B=180^{\circ}{sum of co-interior angles of a parallelogram}
\frac{1}{2}\angle A+\frac{1}{2}\angle B=90^{\circ} {Dividing both sides by 2}
\angle PAD+\angle PDA=90^{\circ}
\Rightarrow \angle APD=90^{\circ} {\because sum of all the angles of a triangle is 1800}
 \angle QRS=90^{\circ}
Similarly,
\angle RBC+\angle RCB=90^{\circ}
\Rightarrow \angle BRC=90^{\circ}  
\angle QRS=90^{\circ}
Similarly,
\angle QAB+\angle QBA=90^{\circ}
\Rightarrow \angle AQB=90^{\circ}  {\because sum of all the angles of a triangle is 1800}
\angle RQP=90^{\circ} {\because vertically opposite angles}
Similarly,|

\angle SDC+\angle SCD=90^{\circ}
\Rightarrow \angle DSC=90^{\circ} {\because sum of all the angles of a triangle is 1800}
\angle RSP=90^{\circ} {\because vertically opposite angles}
Thus PQRS is a quadrilateral whose all angles are 90^{\circ}
Hence PQRS is a rectangle.
Hence proved

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E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF \parallel AB and EF EF=\frac{1}{2}(AB+CD)

Given: ABCD is a trapezium in which AB\parallel CD, E and F are the mid-points or sides of AD and BC.


Constructions: Joint BE and produce it to meet CD at G.
Draw BOD which intersects EF at O
To Prove: EF \parallel AB  and EF=\frac{1}{2}(AB+CD)

Proof: In \triangle GCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem.
EF \parallel GC
But, GC\parallel AB or  CD \parallel AB           {given}
\therefore EF\parallel AB
In \triangle ADB, AB \parallel EO and E is the mid-point of AD. Then by the mid-point theorem, O is the mid-point of BD.
EO=\frac{1}{2}AB          ……(1)
In \triangle BDC, OF \parallel CDand O is the mid-point of BD
OF=\frac{1}{2}CD         …..(2)
Adding 1 and 2, we get
EO+OF=\frac{1}{2}AB+\frac{1}{2}CD
EF=\frac{1}{2}(AB+CD)

Hence Proved

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Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Hint:  Prove all the sides are equal and diagonals are equal.

Given: In a square ABCD; P, Q, R and S are the mid-points of AB, BC, CD and DA.
To Prove: PQRS is a square
Construction: Join AC and BD

Proof: Here ABCD is a square
\ AB = BC = CD = AD
P, Q, R, and S are the mid-points of AB, BC, CD and DA.
In \triangle ADC,
SR\parallel AC
SR=\frac{1}{2}AC  {using mid-point theorem} …..(1)
In \triangle ABC,
PQ\parallel AC
PQ=\frac{1}{2}AC {using mid-point theorem} …..(2)
From equation 1 and 2
SR\parallel PQ  and SR= PQ=\frac{1}{2}AC    …..(3)
Similarly, SP\parallel BD  and BD\parallel RQ
\ SP=\frac{1}{2}BD and  RQ=\frac{1}{2}BD   {using mid-point theorem}
 SP=RQ=\frac{1}{2}BD
Since diagonals of a square bisect each other at right angles.
 AC = BD
\Rightarrow SP=RQ=\frac{1}{2}AC …..(4)
From equation 3 and 4
SP = PQ = SP = RQ {All sides are equal}
In quadrilateral OERF
OE\parallel FR and OF\parallel ER
\angle EOF=\angle ERF=90^{\circ}
In quadrilateral RSPQ
\angle SRQ=\angle ERF=\angle SPQ=90^{\circ}   {Opposite angles in a parallelogram are equal}
\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}
90^{\circ} +90^{\circ}+ \angle RSP + \angle RQP = 360^{\circ}
\angle RSP + \angle RQP = 180^{\circ}
\angle RSP = \angle RQP = 90^{\circ}
Since all the sides are equal and angles are also equal, so PQRS is a square.
Hence Proved

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E is the mid-point of a median AD  \triangle ABC and BE is produced to meet AC at F. Show that AF=\frac{1}{3}AC .

Given: In \triangle ABC, AD is a median and E is the mid-point of AD
To Prove: AF=\frac{1}{3}AC.
Construction: Draw DP\parallel EF and \triangle ABC as given

Proof: In \triangle ADP, E is the mid-point of AD and EF\parallel DB
So, F is the mid-point of AP {converse of midpoint theorem}
In \triangle FBC, D is the mid-point of BC and DP\parallel BF
So, P is the mid-point of FC {converse of midpoint theorem}
Thus, AF = FP = PC
AF+FP+PC=AC=3AF
\therefore AF=\frac{1}{3}AC
Hence Proved

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In Figure, AB \parallel DE, AB = DE, AC \parallel DF and AC = DF. Prove that BC \parallel EF and  BC = EF.

Given: AB \parallel DE  and  AC \parallel DF
Also, AB = DE and AC = DF
To prove:  BC\parallel EFand BC = EF

Proof: AB \parallel DE  and AB = DE
AC \parallel DF and AC = DF
In ACFD quadrilateral
AC\parallel FD and AC = FD   …..(1)
Thus ACFD is a parallelogram
AD\parallel CF and AD = CF …..(2)
In ABED quadrilateral
AB\parallel DE and AB = DE   …..(3)
Thus ABED is a parallelogram
AD\parallel BE  and AD = BE …..(4)
From equation 2 and 4
AD = BE = CF and AD \parallel CF \parallel BE …..(5)
In quadrilateral BCFE, BE = CF and BE\parallel CF   {from equation 5}
So, BCFE is a parallelogram BC = EF and BC\parallel EF
Hence Proved.

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