CHAPTER-8, QUADRILATERAL

Explanation:

 Let us consider the fourth angle as x

 We know that in a quadrilateral, the sum of all the angles is 360°

Therefore, x + 75° + 90° + 75° = 360°

By calculating this, we get:

x = 360° - 75° - 90° - 75°

x = 360° - 240°

x = 120°

Hence proved, the measure of the fourth angle is 120° By subtracting the 3 angles from 360° we get the value of x=120°.

Explanation:

The angle between the side of the rectangle and its diagonal = 25º

Let's consider x as the acute angle between diagonals

As the diagonals of a rectangle are of equal length

AC = BD

Now let's divide LHS and RHS by 2

AC/2 = BD/2

So, here O is the midpoint of AC and BD

Therefore,OD = OC

We know that the angles opposite to equal sides are equal

So, ∠y = 25°

From the angle sum property of a triangle,

∠BOC = ∠ODC + ∠OCD

So by substituting the values, we get

∠x = ∠y + 25°

This implies, ∠x = 25° + 25°

So,∠x = 50°

Hence proved, the acute angle between the diagonals is 50°.

Explanation: 

From the above figure we know that,

∠ACB =∠OCB= 40°

As AD  parallel to BC

∠DAC = ∠BCA = 40° (Alternate interior angles)

∠DAO = 40°

Diagonals of a rhombus are perpendicular to each other, but not equal.

 So, ∠AOD = 90°

From the angle sum property of the triangle, we know  that, 

∠AOD + ∠ADO + ∠DAO = 180°

 By substituting the above  values

90° + ∠ADO + 40° = 180°

130° + ∠ADO = 180°

∠ADO = 180° – 130°

 So,  ∠ADO = 50°

So,  ∠ADB = 50°

Therefore, ∠ADB is 50 degrees.

Explanation:

It is given in the above that PQRS is a quadrilateral.

Here A is the midpoint of the side PQ, B is the midpoint of the side QR, C is the midpoint of the side SR, and D is the mid-point of the side PS.

Here, the quadrilateral ABCD is formed by joining the mid-points of the sides of the quadrilateral and ABCD is a rectangle.

In the triangle QSR, B is the midpoint of QR and C is the midpoint of SR.

Since BC is the line segment that joins the sides QR and SR, So by using the mid-point theorem we have:

BC=1/2SQ

 And also, we have BC parallel to SQ

Therefore, UO parallel to CT…….(a)

By using the mid-point theorem we have,

CD=1/2PR

And also, we have CD parallel to PR

Therefore, UC parallel to OT…….(b)

From (a) and (b), we get a  rectangle UOTC which  is a parallelogram

Since ABCD is a rectangle, ∠C=90°

Also from the definition of a parallelogram, we have ∠C=∠O=90°

 So, we have ∠C=∠O=90°

Since∠O=90°, So we can easily say that PR perpendicular QS

Hence, the diagonals of PQRS are perpendicular.

Explanation:

The  correct answer is (D) 

Here we know that ABCD is a rhombus

So we have,

AB = BC = CD = DA

Here, D and C are midpoints of PQ and PS

So, by the  midpoint theorem,

We have,

DC = ½ QS

Also, B and C are midpoints of SR and PS

So by the  midpoint theorem

We have,

BC = ½ PR

Now we know that ABCD is a rhombus

∴ BC = CD

This implies, ½ QS = ½ PR

This implies, QS = PR

Hence proved, the diagonals of PQRS are equal.

Explanation:

Let's consider the angles of the quadrilateral ABCD to be 3x, 7x, 6x, and 4x.

We all know that the sum of all angles of a quadrilateral is equal to 360 degrees.

So, 3x + 7x + 6x + 4x = 360°

So,10x + 10x = 360°

Therefore,20x = 360°

 Therefore,x = 360°/20

 So, x = 18°

Now, A = 3x = 3(18) = 54°

B = 7x = 7(18) = 126°

C = 6x = 6(18) = 108°

D = 4x = 4(18) = 72°

Her, AD, and BC are two lines cut by a transversal CD.

Now the sum of  ∠C and ∠D on the same side of the transversal,

∠C +∠D =108° + 72° =180°

We all know that the sum of interior angles on the same side of the transversal is equal to 180°, then the two lines are parallel.

So, AD is Parallel to  BC

This implies ABCD is a quadrilateral in which one pair of opposite sides are parallel.

Hence proved, ABCD is a trapezium.

Explanation:

From the angle sum property of a quadrilateral, we all know that the sum of the angles is 360º

So,∠A + ∠B + ∠C + ∠D = 360°

So by dividing both LHS and RHS by 2, we get that

1/2 (∠A + ∠B + ∠C + ∠D) = 1/2 × 360° = 180°

AP, PB, RC, and RD are the bisectors of ∠A, ∠B, ∠C, and ∠D

 So,∠PAB + ∠ABP + ∠RCD + ∠RDC = 180° …. (a)

We know that the sum of all angles of a triangle = 180°

∠PAB + ∠APB + ∠ABP = 180°

∠PAB + ∠ABP = 180° – ∠APB …. (b)

Similarly

∠RDC + ∠RCD + ∠CRD = 180°

∠RDC + ∠RCD = 180° – ∠CRD …. (c )

Let us substitute equations (b) and (c ) in (a)

Here we get that,180° – ∠APB + 180° – ∠CRD = 180°

360° – ∠APB – ∠CRD = 180°

So we get

∠APB + ∠CRD = 360° – 180°

∠APB + ∠CRD = 180° …. (d)

∠SPQ = ∠APB and ∠SRQ = ∠DRC are vertically opposite angles

By substituting  it in equation (d)

We get that,∠SPQ + ∠SRQ = 180°

Therefore, PQRS is a quadrilateral whose opposite angles are supplementary.

Explanation:

Now let's consider that the bisectors of angles APQ and CPQ meet at point M and the bisectors of angles BPQ and PQD meet at point N

Now let's  join PM, MQ, QN and NP

As APB is parallel to CQD,

∠APQ = ∠PQD

As NP and PQ are the angle bisectors

2∠MPQ = 2 ∠NQP

 Now let us divide both sides by 2

∠MPQ = ∠NQP

PM  Parallel to  QN

In the same way,

∠BPQ = ∠CQP

PN  Parallel to  QM

Therefore, PNQM is a parallelogram

We all know that angle on a straight line is 180°

So,∠CQP + ∠CQP = 180°

So,2∠MPQ + 2∠NQP = 180°

Now let's divide both sides by 2

We get that,∠MPQ + ∠NQP = 90°

Therefore,∠MQN = 90°

Hence proved, PMQN is a rectangle.

Therefore, the bisectors of the angles APQ, BPQ, CQP, and PQD form a rectangle.

Explanation:

By the Midpoint Theorem, join AC, RP, and SQ

In triangle ABC,

P is the midpoint of AB and Q is the midpoint of BC

So by  using the midpoint theorem,

PQ  parallel to AC and PQ = 1/2AC …(a)

Similarly,

In triangle DAC,

S is the midpoint of AD and R is the midpoint of CD

So  by  using the midpoint theorem,

SR  parallel to AC and SR = 1/2AC …(b)

From (a) and (b),

PQ  parallel to SR and PQ = SR

this implies PQRS is a parallelogram

ABQS is a parallelogram

this implies AB = SQ

PBCR is a parallelogram

this implies BC = PR

this implies AB = PR [because  BC = AB, sides of rhombus]

this implies SQ = PR

Therefore the diagonals of the parallelogram are equal

Hence Proved, it is a rectangle.

Explanation:

In triangle  ABC

By using  the midpoint theorem

We get, DE  Parallel to  BC

So,DE = 1/2 BC

So we get

DE = 1/2 [BP + PO + OQ + QC]

DE = 1/2 [2PO + 2OQ]

As P and Q are the midpoints of OB and OC

So, DE = PO + OQ

Therefore DE = PQ

In triangle  AOC

Q and C are the midpoints of AC and OC

In triangle  AOB

PD  Parallel to  AO

PD = 1/2 AO [ by using mid-point theorem]

From triangle  AOC and triangle  AOB

EQ  Parallel to  PD and EQ = PD

From triangle  ABC

DE  Parallel to  BC or DE  Parallel to  PQ

DE = PQ

Hence proved, DEQP is a parallelogram

Therefore, if P and Q are the mid-points of OB and OC respectively, then DEQP is a parallelogram.

Explanation:

From the above figure we know that,

ABCD is a parallelogram

P, Q, R and S are the midpoints of AB, BC, CD, and DA

So PQRS is a square

PQ = QR = RS = PS … (a)

So,PR = SQ

As PR = BC and SQ = AB

So,AB = BC

Therefore all sides of quadrilateral ABCD are equal

By using the midpoint theorem

SP  Parallel to  DB, SP = 1/2 DB … (b)

In triangle  ABC,

PQ  Parallel to  AC, PQ = 1/2 AC … (c)

From equation (a)

PS = PQ

From equation (b) and (c)

1/2 DB = 1/2 AC

DB = AC

Therefore, ABCD is a square as the diagonals are equal And here the diagonals are perpendicular.

Therefore, the figure is a square only if the diagonals of ABCD are equal and perpendicular.

Explanation:

It is given in the above that,

ABCD is a parallelogram

 And AC and BD are the diagonals

So, AD is Parallel to  BC

Therefore ∠DAC = ∠ACB [Alternate angle]

So,∠ACB = 32º

Here we know that

∠AOB + ∠BOC = 180º [Straight line]

By substituting the values,

70º + ∠BOC = 180º

So,∠BOC = 110º

In triangle  BOC,

∠OBC + ∠BOC + ∠OCB = 180º

By substituting the values

∠OBC + 110º + 32º = 180º

So,∠OBC = 38º

Therefore,∠DBC = 38º.

Explanation:

According to Euclidean geometry, “A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

By considering the properties of a parallelogram, the above statement “opposite angles are bisected by the diagonals” is false.

Explanation:

In ∆ADE and ∆CFE,

Let us assume that, DE = EF

And  E is the midpoint of AC

So,AE = CE

If DE = EF,

Then ∠AED = ∠FEC [vertically opposite angles]

By using  the SAS congruence rule

∆ADE ≅ ∆CFE

 So,AD = CF [ by c.p.c.t]

∠ADE = ∠CFE [ by c.p.c.t]

We know that alternate angles are equal

Therefore, AD  Parallel to  CF

Therefore, we need additional information which is DE = EF.

Explanation:

From the above diagram, we know that, AB = CD and AD = BC

In ∆ABD and ∆ACB

AD = BC

BD = AC

AB = AB

From the SSS criterion,

∆ABD ≅ ∆ACB

So,∠BAD = ∠ABC

We know that AD is Parallel to  BC and the transversal intersects them at A and B

Therefore,∠BAD + ∠ABC = 180º

 So, ∠ABC + ∠ABC = 180º

So, 2∠ABC = 180º

Now let's divide both sides by 2

Therefore,  ∠ABC = 90º

Hence proved, the value of ∠ABC is 90º

Explanation:

In a rectangle, we know that

AB = CD and AD = BC

Here

AC² = AD² + CD²

If AD = BC

Then,AC² = BC² + CD²

So,AC² = BD²

By taking square root on both the sides

AC = BD

so, the diagonals AC and BD are equal to each other but we cannot consider them to be perpendicular

Therefore, the above statement is false.

Explanation:

Let ∠A = ∠B = ∠C = ∠D = 70º

We all know that the sum of all the angles of a parallelogram is 360º

Therefore, ∠A + ∠B + ∠C + ∠D = 360º

By substituting the values,

70º + 70º + 70º + ∠D = 360º

so, by further calculation, we get that,

∠D = 360º - 210º

Therefore, ∠D = 150º

Because here the opposite angles are not equal and the consecutive angles are not supplementary

Hence proved, ABCD is not a parallelogram.

Explanation:

By Considering the properties of parallelogram

As OA: OC = 3: 2 is not equal

Hence proved, ABCD is not a parallelogram

Therefore, OA is not equal to OC.

Explanation:

 Given, OA=3cm and OD=2 cm

We know that diagonals of a parallelogram bisect each other.

therefore,  AO=OC and BO=OD

This implies  AC=2×OA=2×3cm=6cm

This implies,  BD=2×OD=2×2cm=4cm

Therefore the length of the diagonals AC and BD are 6cm and 4cm.

Explanation:

By considering the above diagram, AC and BD are the diagonals And they intersect each other at point O By considering the properties of a parallelogram the Diagonals of a parallelogram bisect each other but not at 90°. So the diagonals are not perpendicular

Therefore, the above statement is false.

Explanation:

 The sum of all the angles of a quadrilateral is 360º

So the sum of the given angles is

110º + 80º + 70º + 95º = 355º ≠ 360º

So , 110º, 80º, 70º, and 95º cannot be the angles of a quadrilateral.

Explanation:

In quadrilateral ABCD, ∠A + ∠D = 180º,

In a trapezium, we know that

The sum of co-interior angles is 180°

Therefore it is proved that the given quadrilateral is a trapezium.

Explanation:

If all the angles of the quadrilateral = x

We know that, Sum of all angles of a quadrilateral = 360°

This implies, x + x + x + x = 360°

so,4x = 360°

so, x = 360°/4

therefore, x = 90°

Therefore it is proved that the quadrilateral is a rectangle.

Explanation:

According to the properties of a rectangle, the diagonals of a rectangle are equal and bisect each other as shown in the figure. So, they are equal but they are not perpendicular.

Hence proved, the above-given statement is false.

Explanation:

An obtuse angle is greater than 90º. According to the angle sum property, the sum of all four angles of a quadrilateral is 360º

So If we consider all four angles more than 90º, the sum will be more than 360º

So all four angles cannot be obtuse in a quadrilateral.

Hence proved, all four angles cannot be obtuse.

Explanation:

It is given in the above  that, in triangle ABC  and

AB = 5 cm

BC = 8 cm

CA = 7 cm

Here, D and E are respectively the mid-points of AB and BC

Now we have to find the length of DE

By using the mid-point theorem, we get that

DE = 1/2 AC

By substituting the values,

DE = 1/2 (7)

so, DE = 3.5 cm

Therefore, the length of DE is 3.5 cm.

Explanation:

ABC is a triangle, and

D, E, and F are the points on BC, CA, and AB

here,BDEF and FDCE are parallelograms

In parallelogram BDEF,

BD = EF …. (a) [because  the opposite sides of a parallelogram are equal]

In parallelogram FDCE,

CD = EF …. (b) [because the opposite sides of a parallelogram are equal]

From the equations (a) and (b)

We get that, BD = CD

Hence, it is proved that BD = CD.

Explanation:

Here, AD or AG and AB or AE are the two sides that are sharing

∠C = 55º

Because we know that the opposite angles of a parallelogram are equal

so,∠F = ∠A

∠A = ∠C

So we get

∠F = ∠C

As ∠C = 55º, ∠F = 55º

Hence , ∠F = 55º.

Explanation:

We know that an acute angle is less than 90º. So all the angles of a quadrilateral cannot be acute angles because the sum of angles of a quadrilateral will be less than 360º, whereas the angle sum of a quadrilateral is 360º

Explanation:

If all the angles of a quadrilateral is a right angles, then it will be

90º + 90º + 90º + 90º = 360º

According to  the angle sum property of a quadrilateral, the sum of all angles is 360º

So the quadrilateral will be a rectangle or a square.

Hence proved, the angles can be right angles.

Explanation:

It is given that the diagonals of a quadrilateral ABCD bisect each other, so it is a parallelogram

Given, ∠A = 35º

We know that the sum of interior angles between two parallel lines is 180º

So, ∠A + ∠B = 180º

By substituting the values, we get that

35º + ∠B = 180º

So, ∠B = 180º - 35º

So we get

∠B = 145º

Therefore, ∠B = 145º.

Explanation:

Opposite sides of the quadrilateral ABCD are equal.

This implies AB=CD

It is Given in the above that, AB=4 cm

Therefore, CD=4 cm.

Explanation:

We know that the sum of all the angles of a quadrilateral is equal to 360 degrees.

Let us consider the angles of the quadrilateral to be 3x, 4x, 4x, and 7x.

So, 3x + 4x + 4x + 7x = 360º

so,10x + 8x = 360º

therefore,18x = 360º

x = 360º/18

x = 60º/3

x = 20º

Now, 3x = 3(20º) = 60º

And 4x = 4(20º) = 80º

And 7x = 7(20º) = 140º

Hence, the angles of the quadrilateral are 60º, 80º, 80º, and 140º.

Explanation:

We know that opposite sides of a parallelogram are equal.

So, AD = BC ----------- (a)

Since we know that  X is the midpoint of AD

AX = DX

So, AD = AX + DX

AD = DX + DX

i,e,AD = 2DX --------------- (b)

Since Y is the midpoint of BC

BY = CY

So, BC = BY + CY

BC = BY + BY

i.e,BC = 2BY ------------------ (c)

Using (b) and (c) in (a),

2DX = 2BY

DX = BY

We know that a pair of opposite sides are equal and parallel in a parallelogram.

So, DX is Parallel to  BY

This implies XBYD is a parallelogram

So, PX  Parallel to  QD

From triangle AQD,

Since X is the midpoint of AD

therefore,AP = PQ ------------ (d)

Similarly, from triangle CPB

CQ = PQ ------------ (e)

From (d) and (e),

AP = PQ = CQ

Therefore, AP = PQ = QC.

Explanation:

It is given in the above that, ABCD is a parallelogram,

∠DAB = 2x

∠DCB = 2y

Since we know that opposite angles of a parallelogram are equal.

So, ∠A = ∠C

so,2x = 2y

i.e,x = y

As DC is Parallel to  AB and XC is Parallel to  AY

∠XCY = ∠CYB [Alternate angles]

so,∠CYB = x

∠XAY = x

As ∠XAY and ∠CYB are corresponding angles

Therefore, AX is Parallel to  CY

Hence proved, AX is parallel to CY.

Explanation:

Let us consider the remaining three equal angles to be x.

We know that the Sum of all interior angles of a quadrilateral is = 360 degree

Therefore, 108º + x + x + x = 360º

So, 108º + 3x = 360º

This implies,3x = 360º – 108º

This implies,3x = 252º

Therefore, x = 252º/3

Therefore, x = 84º

So that, each of three equal angles, x = 84º.

Explanation:

As we know that opposite angles in a quadrilateral are always supplementary.

So, ∠B + ∠C = 180º

It is already given, ∠B = 45º

So, 45 + ∠C = 180º

∠C = 180º - 45º

Therefore, ∠C = 135º

Similarly, ∠A + ∠D = 180º

It is already given, ∠A = 45º

Therefore, 45º + ∠D = 180º

∠D = 180º - 45º

Therefore  ,∠D = 135º

Hence proved, the angles C and D of the trapezium are equal to 135º.

Explanation:

Let's consider, ABCD is a parallelogram

And, ∠ADC and ∠ABC are the two obtuse angles of the parallelogram

Here DQ and DP are the two altitudes of the parallelogram

So,  DP perpendicular  AB

And DQ perpendicular  BC.

Therefore, ∠PDQ = 60

In quadrilateral DPBQ,

As we know the quadrilateral angle sum property states that, the sum of all four interior angles is 360º.

The Sum of all interior angles of a quadrilateral is = 360º

Here we have,

∠PDQ + ∠Q + ∠P + ∠B = 360º

I.e, 60 + 90 + 90 + ∠B = 360º

240 + ∠B = 360º

Therefore, ∠B = 360º - 240º

Therefore ,∠B = 120º

Since the  opposite angles in a parallelogram are equal,

∠B = ∠D = 120º

Since the opposite sides are parallel in a parallelogram,

AB Parallel to CD

And also, since the sum of adjacent interior angles is 180º

∠B + ∠C = 180º

120 + ∠C = 180º

so,∠C = 180º - 120º

I.e, ∠C = 60º

Since we know that the opposite angles in a parallelogram are equal,

∠C = ∠A = 60º

Therefore, areas of the parallelogram are 60º, 120º, 60º and 120º

Explanation:

As mentioned above we know that ABCD is a rhombus.

And DE is the altitude on AB then AE = EB.

In triangle AED and triangle BED,

We have,

DE = DE (as it is a common line)

And, ∠AED = ∠BED (right angle)

AE = EB (DE is an altitude)

Therefore, triangle AED ≅ triangle BED by SAS property. 

Therefore,  AD = BD (by C.P.C.T)

But AD = AB (sides of a rhombus are equal)

This implies, AD = AB = BD

So ABD is an equilateral triangle.

∴ ∠A = 60º

As we know that the opposite angles of a rhombus are equal, we get,

this implies ∠A = ∠C = 60º

We also know that, the sum of adjacent angles of a rhombus = supplementary.

So,

∠ABC + ∠BCD = 180º

∠ABC + 60o = 180º

∠ABC = 180º – 60º= 120º

Since the opposite angles of a rhombus are equal, here  we get that,

∠ABC = ∠ADC = 120º

Hence, the Angles of a rhombus are:

∠A = 60º, ∠C = 60º, ∠B = 120º, ∠D = 120º.

Explanation:

Join BD, and meet AC at point O.

Since diagonals of a parallelogram bisect each other.

Therefore,  OA=OC

and OD=OB

Now, OA=OC

and AE=CF               [ Given ]

This implies,  OA−AE=OC−CF

So,  OE=OF

Thus, BFDE is a quadrilateral whose diagonals bisect each other.

Hence Proved  BFDE is a parallelogram.

Explanation:

From the above-mentioned point in question

Let EF intersect BD at G.

In triangle ABD,

E is the midpoint of AD and also EG is parallel to AB.

Here we get, G is the midpoint of BD             [ By converse of midpoint theorem ]

Similarly,

In triangle BDC,

G is the midpoint of BD and GF is parallel to AB parallel to DC.

Therefore,  F is the midpoint of BC          [ By converse of midpoint theorem ]

Explanation:

As it is given, RQ  Parallel to  BC

PR  Parallel to  AC

And QP  Parallel to  AB

By considering the quadrilateral BCAR,

BR  Parallel to  CA

And RA  Parallel to  BC

As we know, the opposite sides of a parallelogram are parallel and congruent.

So, BCAR is a parallelogram.

BC = AR ------------------ (a)

Considering quadrilateral BCQA,

BC  Parallel to  AQ

And AB  Parallel to  QC

So, BCQA is a parallelogram

BC = AQ ------------------ (b)

Adding (a) and (b),

BC + BC = AR + AQ

So, BC = AR + AQ

From the figure,

AR + AQ = RQ

So, 2BC = RQ

Therefore, BC = 1/2 QR.

Explanation:

Let ABC be the triangle and D, E, and F be the midpoint of BC, CA, and AB respectively. As we know, the line segment joining the mid-points of two sides of a triangle is half of the third side.

Therefore DE= 1/2AB,EF= 1/2 BC and FD= 1/2 AC

Now, triangle ABC is an equilateral triangle 

this impliesAB=BC=CA

this implies 1/2 AB= 1/2 BC= 1/2 CA

So, DE=EF=FD

Hence proved, triangle DEF is an equilateral triangle.

Explanation:

In triangle APO and  triangle OQC

∠OAP=∠OCQ ( taking AC as transversal then angles are on opposites sides of transversal)

      AP=CQ  (as given in the above )

∠OPA=∠OQC ( taking PQ as transversal then angles are on opposites sides of transversal )

By the ASA criterion of congruence, triangle APO is congruent to triangle OQC.

Therefore, PO=OQ, AO=OC ( Corresponding parts of a congruent triangle are equal )

So, AC and PQ bisect each other.

Explanation:

So, According to the given data 

∠BAP=∠BAP

∠BPA=∠BAP

So, AB=BP

and AB∣∣CD 

So, AD=BC

and P is the midpoint of BC

So, 2BP=BC

i.e,2BP=AD

i.e, AD=2AB

So, ​AB=CD

Therefore, AD=2CD.

Explanation:

As we know that the opposite sides of a parallelogram are parallel and congruent.

So, PQ = RS and PQ Parallel to  RS --------- (a)

Here ,PQRS is a parallelogram.

We also know that the sum of interior angles lying on the same side of the transversal is always supplementary.

∠RPQ + ∠PQS = 180°

Now, ∠RPQ + ∠PQM + ∠MQS = 180° --------- (b)

Also, PN  Parallel to  QM

So, ∠NPQ + ∠PQM = 180°

Now, ∠NPR + ∠RPQ + ∠PQM = 180° ----------- (c)

By comparing (b) and (c),

We get that, ∠MQS = ∠NPR ------------- (d)

Similarly, ∠MSQ = ∠NRP ---------- (e)

From (a), (d), and (e)

According to  ASA criteria, the triangles PNR and QMS are congruent.

And by CPCTC,

NR = MS

PN = QM

So, PN  Parallel to  QM

Therefore, PQMN is a parallelogram

As we know that the opposite sides of a parallelogram are parallel and congruent.

MN = PQ

NM  Parallel to  PQ

Hence it is proven that MN and PQ are equal and parallel to each other.

Explanation:

Here the diagonal AC divides the parallelogram ABCD into two triangles ABC and ADC.

Now by considering triangles ABC and ADC,

We know that the opposite sides of a parallelogram are parallel and congruent.

So, AD = BC

And, AD Parallel to BC

As we know that the alternate interior angles are equal.

Therefore, ∠BAC = ∠DCA

And ∠BCA = ∠DAC

AC is the common side

Now here we can observe that one side and two angles made on these sides are equal.

According to  ASA criteria, the triangles ABC and ADC are similar.

Therefore, triangle ABC ≅ triangle ADC.

Explanation:

Let's consider a rhombus ABCD

Here the points P, Q, R, and S are the midpoints of the sides AB, BC, CD and AD.

Now we have to show that PQRS is a rectangle.

Let's join the diagonals AC and BD of the rhombus ABCD.

Now by considering the triangle ABD,

Since S and P are the midpoints of the sides AD and AB.

SP  Parallel to  BD ----------------- (a)

SP = 1/2 BD --------------- (b)

Similarly, RQ  Parallel to  BD

RQ = 1/2 BD -------------- (c)

From (b) and (c),

SP = RQ

Also, SP  Parallel to  RQ

Therefore, PQRS is a parallelogram

As we know that the diagonals of a rhombus are perpendicular.

So, AC perpendicular  BD -------------- (d)

Considering triangle BAC,

PQ  Parallel to  AC --------------- (e)

From (a), (d), and (e),

SP perpendicular  PQ

i.e.,∠SPQ = 90°

As we know, a rectangle is a quadrilateral with four right angles. The opposite sides are parallel and equal to each other.

Therefore, PQRS is a rectangle.

Explanation:

Let's consider a parallelogram ABCD

Since AC bisects angle A

∠CAB = ∠CAD ------------------ (a)

As we know that the opposite sides of a parallelogram are parallel and congruent.

So, AB Parallel to CD and AC is a transversal.

As we know that the alternate interior angles are equal.

Therefore,  ∠CAB = ∠ACD ------------- (b)

Similarly, AD Parallel to BC and AC will be transversal.

Therefore  ,∠DAC = ∠ACB -------------- (c)

From (a), (b), and (c),

We get that,∠BCA = ∠DCA

This implies AC bisects the opposite angle C.

Hence proved that AC bisects its opposite angle also.

Explanation:

Let's consider an isosceles right triangle ABC right angled at A.

A square DEF is inscribed in the triangle.

Given, ∠A = 90°

As we know, AB=AC --------------- (a)

As we know that all sides of a square are equal

So, AD = AF ---------------- (b)

By subtracting (a) and (b), we get

AB - AD = AC - AF

Therefore,BD = CF -------------------- (c)

By  considering triangles CFE and BDE,

Therefore the sides of a square DE = EF

From (c), BD = CF

Therefore,∠CFE = ∠EDB = 90°

By SAS criteria, the triangles CFE and BDE are similar.

So by CPCTC,

CE = BE

Hence proved, vertex E of the square bisects the hypotenuse BC.

Explanation:

,

As mentioned above, ABCD is a parallelogram

Since AF bisects ∠A,

We get,

∠BAE = ∠EAD … (a)

∠EAD = ∠EFB … (b) [Alternate angles]

So from the above equations (a) and (b),

We get that,

∠BAE = ∠EFB

Since we know that the sides opposite to equal angles are equal,

BF = AB

Here, AB = 10 cm

So, BF = 10 cm

This implies, BC + CF = 10 cm

6 cm + CF = 10 cm [BC = AD = 6 cm, opposite sides of a parallelogram]

This implies, CF = 10 – 6 cm = 4 cm

This implies, CF = 4 cm.

Explanation:

Here we have,

P is the midpoint of the sides AB

Q is the midpoint of the sides of BC

R is the midpoint of the sides CD

S is the midpoint of the sides DA

Here also, we know that,

AC = BD.

In triangle ADC, by mid-point theorem,

Therefore, SR = 1/ 2 AC

And, SR Parallel to AC

In triangle ABC, by mid-point theorem,

Therefore ,PQ = 1/2 AC

And, PQ Parallel to AC

Hence, SR = PQ = 1/2 AC

Similarly,

In triangle BCD, by mid-point theorem,

Therefore ,RQ = 1/2 BD

And, RQ Parallel to BD

In triangle BAD, by mid-point theorem,

Therefore ,  SP = 1/2 BD

And, SP Parallel to BD

So, we get,

SP = RQ = 1/2 BD = 1/2 AC

Then,

SR = PQ = SP = RQ

Hence proved that PQRS is a rhombus.

Explanation:

As mentioned above, we know that, AC perpendicular  BD

Therefore ,∠AOD = ∠AOB = ∠BOC = ∠COD = 90°

By considering triangle ADC,

S and R are the midpoints of AD and DC

By the midpoint theorem,

SR  Parallel to  AC

Therefore ,SR = 1/2 AC --------------- (a)

By considering triangle ABC,

P and Q are the midpoints of AB and BC

By the midpoint theorem,

PQ  Parallel to  AC

Therefore,PQ = 1/2 AC ----------------- (b)

So by comparing (a) and (b)

SR = PQ = 1/2 AC ------------ (c)

By considering triangle BAD,

SP  Parallel to  BD

By the midpoint theorem,

Therefore ,SP = 1/2 BD ----------------- (d)

Comparing (c) and (d),

SP = RQ = 1/2 BD ----------- (e)

Considering quadrilateral EOFR,

OE  Parallel to  FR

OF  Parallel to  ER

Therefore ,∠EOF = ∠ERF = 90°

Hence proving that PQRS is a rectangle.

Explanation:

By considering the above  question,

We have,

P is the mid-point of the sides AB

Q is the midpoint of the sides of BC

R is the midpoint of the sides CD

S is the midpoint of the sides DA

Also, it is given that

AC perpendicular  BD

And AC = BD

In triangle ADC, by mid-point theorem,

SR = 1/2 AC

And, SR Parallel to AC

In triangle ABC, by mid-point theorem,

PQ = 1/2 AC

And, PQ Parallel to AC

So, we have,

PO Parallel to SR and PQ = SR = 1/2 AC

Now, in triangle ABD, by mid-point theorem,

SP Parallel to BD and SP = 1/2 BD = 1/2AC

In triangle BCD, by mid-point theorem,

RQ Parallel to BD and RQ = 1/2 BD = ½1/2AC

Therefore ,SP = RQ = 1/2 AC

Therefore,PQ = SR = SP = RQ

Thus, we get that,

Therefore all four sides are equal.

Here by considering the quadrilateral EOFR,

 We get that, OE Parallel to FR, OF Parallel to ER

∠EOF = ∠ERF = 90o (because opposite angles of the parallelogram)

∠QRS = 90°

Hence proved, PQRS is a square.

Explanation:

Since AC bisects angle A

∠CAB = ∠CAD ------------------ (a)

As we know that the opposite sides of a parallelogram are parallel and congruent.

So, AB Parallel to CD and AC is a transversal.

As we know that the alternate interior angles are equal.

∠CAB = ∠ACD ----------------- (b)

Similarly, AD Parallel to BC and AC is a transversal.

∠DAC = ∠ACB ----------------- (c)

From (a), (b), and (c),

We get that,∠BCA = ∠DCA

As we know that the opposite angles of a parallelogram are equal.

Therefore, ∠A = ∠C

By dividing by 2 into both sides,

1/2 ∠A = 1/2 ∠C

By comparing (a) and (b),

We get that, ∠CAD = ∠ACD

As we know that the sides opposite to the equal angles are equal.

So , CD = AD

As we know that the opposite sides of a parallelogram are parallel and congruent.

Therefore ,AB = CD

And AD = BC

So, AB = BC = CD = AD

This implies all sides are equal. Hence proving ABCD is a rhombus.

Explanation:

As mentioned the above  P is the midpoint of AB

So ,AP = PB

AB = AP + PB

AB = AP + AP

AB = 2AP

Therefore ,AP = 1/2 AB ----------- (a)

As mentioned the above  Q is the midpoint of the CD

So ,QC = QD

CD = QC + QD

CD = QC + QC

CD = 2QC

Therefore ,QC = 1/2 CD ------------ (b)

As we know that the opposite sides of a parallelogram are parallel and congruent

Therefore, AB  Parallel to  CD

Also, AB = CD

By dividing by 2 into both sides,

1/2 AB = 1/2 CD

From (a) and (b),

We get that, AP = QC

Also, AP  Parallel to  QC

Therefore, APCQ is a parallelogram

So, AQ  Parallel to  PC or SQ  Parallel to  PR

Similarly, AB  Parallel to  DC or BP  Parallel to  DQ

So, AB = DC

By dividing by 2 into both sides,

1/2 AB = 1/2 DC

Since P is the midpoint of AB

AP = PB

AB = AP + PB

AB = BP + BP

AB = 2BP

Therefore ,BP = 1/2 AB ----------- (c)

Since Q is the midpoint of the CD

QC = QD

CD = QC + QD

CD = QD + QD

CD = 2QD

Therefore ,QD = 1/2 CD ------------ (d)

From (c) and (d),

BP = QD

Therefore, BPDQ is a parallelogram

As we know that the opposite sides of a parallelogram are parallel and congruent

PD  Parallel to  BQ

Similarly, PS  Parallel to  QR

So, SQ  Parallel to  RP

Therefore, PS  Parallel to  QR

Hence proved, PQRS is a parallelogram.

Explanation:

Let's draw DP perpendicular  AB and CQ perpendicular  AB.

Proof: In triangle APD and triangle BQC,

Since ∠1 and ∠2 are equal to 90°

∠1 = ∠2

Distance between parallel lines,

AB = BC [Given]

According to the  RHS criterion of congruence,

Here we have

triangle APD ≅ triangle BQC [CPCT]

∠A = ∠B

Now, DC Parallel to AB

Since the sum of consecutive interior angles is 180°

∠A+∠3 =180° …(a)

And,

∠B +∠4 =180° …(b)

From equations (a) and (b),

We get that,

∠A + ∠3 = ∠B + ∠4

Since, ∠A = ∠B,

Here we have,

This implies, ∠3 = ∠4

This implies, ∠C = ∠D

Hence, proved, ∠A = ∠B and ∠C = ∠D.

Explanation:

By considering quadrilateral ABED,

As given, AB  Parallel to  DE

AB = DE

So, ABED is a parallelogram

Now, AD  Parallel to  BE

AD = BE ------------------------- (a)

By considering quadrilateral ACFD,

AC  Parallel to  FD

AC = FD

So, ACFD is a parallelogram

Now, AD  Parallel to  FC

AD = FC -------------------------- (b)

From (a) and (b),

AD = BE = FC

CF  Parallel to  BE

So, BCFE is a parallelogram.

As we know BC = EF

BC  Parallel to  EF

Therefore, it is proven that BC = EF, and BC is Parallel to  EF.

Explanation:

As mentioned above, ABC is a triangle

Draw a line  DP parallel to EF in the above diagram 

Considering triangle ADP,

E is the midpoint of the AD line 

EF  Parallel to  DP

By converse of the midpoint theorem,

F is the midpoint of AP.

Considering triangle FBC,

D is the midpoint of BC

DP  Parallel to  BF

By converse of the midpoint theorem,

P is the midpoint of FC

So, AF = FP = PC

Therefore, AF = 1/3 AC.

Explanation:

With a square ABCD

P, Q, R, and S are the midpoints AB, BC, CD, and DA

We have to show that PQRS is a square.

Join the diagonals AC and BD of the square ABCD

We know that all the sides of the square are equal in length

So, AB = BC = CD = AD

Considering triangle ADC,

S and R are the midpoints of AD and DC

By the midpoint theorem,

SR  Parallel to  AC

SR = 1/2 AC --------------- (1)

Considering triangle ABC,

P and Q are the midpoints of AB and BC

By the midpoint theorem,

PQ  Parallel to  AC

PQ = 1/2 AC ----------------- (2)

Comparing (1) and (2),

SR = PQ = 1/2 AC ------------ (3)

Considering triangle BAD,

SP  Parallel to  BD

By the midpoint theorem,

SP = 1/2 BD ----------------- (5)

Comparing (4) and (5),

SP = RQ = 1/2 BD ----------- (6)

We know that the diagonals of a square bisect each other at a right angle.

So, AC = BD

Dividing by 2 into both sides,

1/2 AC = 1/2 BD

From (3) and (6),

SR = PQ = SP = RQ

Considering quadrilateral OERF,

OE  Parallel to  FR

OF  Parallel to  ER

∠FOE = ∠FRE = 90°

Therefore, PQRS is a square.

Explanation:

Consider the above fig in  ABCD 

Join BE and extend it to meet CD produced at G

Draw BD which intersects EF at O.

Consider triangle GCB,

E and F are the midpoints of BG and BC.

By the Midpoint theorem,

EF  Parallel to  GC

Given, AB  Parallel to  GC or CD  Parallel to  AB

So, EF is Parallel to  AB

Considering triangle ABD,

AB  Parallel to  EO

E is the midpoint of AD

By converse of the midpoint theorem,

O is the midpoint of BD

EO = 1/2 AB ------------ (1)

Considering triangle BDC,

OF  Parallel to  CD

O is the midpoint of BD

By converse of the midpoint theorem,

OF = 1/2 CD ----------- (2)

Adding (1) and (2),

EO + OF = 1/2 AB + 1/2 CD

From the figure,

EF = EO + OF

Therefore, EF = 1/2 (AB + CD).

Explanation:

ABCD is a parallelogram 

∠A + ∠D = 180° -------------- (1)

Dividing by 2 into both sides,

1/2 ∠A + 1/2 ∠D = 180°/2

=> 1/2 ∠A+ 1/2 ∠D = 90°

∠PAD + ∠PDA = 90° --------- (2)

Considering triangle PDA,

By angle sum property,

∠APD + ∠PAD + ∠PDA = 180°

From (2),

∠APD + 90° = 180°

∠APD = 180° - 90°

∠APD = 90°

We know  ∠APD = ∠SPQ

∠SPQ = 90°

Similarly, ∠PQR = 90°

∠QRS = 90°

∠PSR = 90°

PQRS is a quadrilateral with each angle equal to 90°.

Therefore, PQRS is a rectangle.

Explanation:

ABCD is a parallelogram

Considering triangles ODP and OBQ,

We know that the vertically opposite angles are equal.

∠BOQ = ∠POD

We know that the alternate interior angles are equal.

∠OBQ = ∠ODP

Given, OB = OD

By ASA criteria, the triangles ODP and OBQ are congruent.

By CPCTC rule,

OP = OQ

Therefore, PQ is bisected at O.

Explanation:

In the Above fig  ABCD is a rectangle

Join the diagonal line AC

Considering triangles BAD and BCD,

Since BD is the bisector of ∠B

∠ABD = ∠CBD

∠A = ∠C = 90°

Common side = BD

By ASA criteria, the triangles BAD and BCD are congruent.

By CPCTC,

AB = BC and AD = CD

We know AB = CD and BC = AD

So, AB = BC = CD = AD

Therefore, ABCD is a square.

Explanation:

In the Below figure, ABC is a triangle

Since D is the midpoint of AB

AD = DB = 1/2 AB

Since E is the midpoint of BC

BE = EC = 1/2 BC

Since F is the midpoint of AC

AF = FC = 1/2 AC

By the midpoint theorem,

EF  Parallel to  AB

EF = 1/2 AB

So, EF = AD = BD

Also, ED  Parallel to  AC

ED = 1/2 AC

So, ED = AF = FC

Similarly, DF  Parallel to  BC

DF = 1/2 BC

So, DF = BE = CE

Considering triangles ADF and EFD,

AD = EF and AF = DE

Common side = FD

By the SSS criterion, the triangles ADF and EFD are congruent.

Similarly,

triangle DEF ≅triangle EDB.

triangle DEF ≅triangle CFE .

Therefore, the triangle ABC is divided into four congruent triangles.

Explanation:

In the above diagram, ABCD is a trapezium

Join CN and extend it to meet AB at E.

Considering triangles CDN and EBN,

Since N is the midpoint of BD

DN = BN

We know 

∠DCN = ∠NEB

∠CDN = ∠NBE

By ASA criterion, the triangle CDN ≅ triangle EBN.

By CPCTC,

DC = EB

CN = NE

Considering triangle CAE,

M and N are the midpoints of AC and CE

MN  Parallel to  AE

By the midpoint theorem,

MN  Parallel to  AB  Parallel to  CD

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

Explanation:

In the above figure, ABCD is a parallelogram

BC  Parallel to  AD and BC = AD

AB  Parallel to  CD and AB = CD

Since P is the midpoint of DC.

DP = PC = 1/2 CD ----------- (1)

Given, QC  Parallel to  AP

PC  Parallel to  AQ

Therefore, APCQ is a parallelogram

So, AQ = PC

From (1),

AQ = PC = 1/2 CD

Since AB = CD

PC = 1/2 AB = BQ

Considering triangles AQR and BQC,

AQ = BQ

We know ∠AQR = ∠BQC

We know ∠ARQ = ∠BCQ

By ASA, the triangles AQR and BQC

By CPCTC,

AR = BC

Given, BC = AD

So, AR = AD

By CPCTC,

CQ =QR

Hence proved that AR = AD and CQ = QR.