Quadrilaterals

Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively

As the sum of all interior angles of a quadrilateral is 360⁰,

3x + 5x + 9x + 13x = 360⁰

30x = 360⁰

x = 12⁰

Hence, the angles are

3x = 3 × 12 = 36⁰
5x = 5 × 12 = 60⁰

9x = 9 × 12 = 108⁰
13x = 13 × 12 = 156⁰

Let ABCD be a parallelogram. To show that ABCD is a rectangle,

We have to prove that: One of its interior angles is 90⁰

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given that the diagonals are equal)

ΔABC ΔDCB (By SSS Congruence rule)

∠ABC = ∠DCB

It is known that the sum of the measures of angles on the same side of transversal is 180⁰

∠ABC + ∠DCB = 180⁰ (AB || CD)

∠ABC + ∠ABC = 180⁰

∠ABC = 90⁰

Since ABCD is a parallelogram and one of its interior angles is 90⁰, ABCD is a rectangle.

To Prove: If diagonals of a quadrilateral bisect at 90º, it is a rhombus.

Figure:

Definition of Rhombus: A parallelogram whose all sides are equal.
Given: Let ABCD be a quadrilateral whose diagonals bisect at 90º

In ΔAOD and ΔCOD,

OA = OC (Diagonals bisect each other)

∠AOD = ∠COD (Given)

OD = OD (Common)

ΔAOD ΔCOD (By SAS congruence rule)

AD = CD ..................(1)

Similarly,

AD = AB and CD = BC ..................(2)

From equations (1) and (2),

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that

ABCD is a rhombus

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.

To prove that the diagonals of a square are equal and bisect each other at right angles,

We have to prove: AC = BD, OA = OC, OB = OD and ∠AOB = 90⁰

In ΔABC and ΔDCB,

AB = DC (Sides of a square are equal to each other)

∠ABC = ∠DCB (angles of square are 90°)

BC = CB (Common side)
ΔABC ΔDCB (By SAS congruency)

AC = DB (By CPCT)

Hence, the diagonals of a square are equal in length.

In ΔAOB and ΔCOD,

∠AOB =∠ COD (Vertically opposite angles)

∠ABO = ∠CDO (Alternate interior angles)

AB = CD (Sides of a square are always equal)

ΔAOB ΔCOD (By AAS congruence rule)

AO = CO and

OB = OD (By CPCT)

Hence, the diagonals of a square bisect each other

In ΔAOB and ΔCOB,

As we had proved that diagonals bisect each other, therefore,

AO = CO

AB = CB (Sides of a square are equal)

BO = BO (Common)

ΔAOB ΔCOB (By SSS congruency)

∠AOB =∠ COB (By CPCT)

However,

∠ AOB + ∠COB = 180⁰ (Linear pair)

2∠AOB = 180⁰

∠AOB = 90⁰

Hence, the diagonals of a square bisect each other at right angles

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.

It is given that the diagonals of ABCD are equal and bisect each other at right angles.

Therefore, AC = BD, OA = OC, OB = OD, and

∠AOB = ∠BOC = ∠COD = ∠AOD = 90⁰

To prove: ABCD is a square,
Proof:

we have to prove that ABCD is a parallelogram with all of its sides equal and one of the interior angle is right angle.

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

∠AOB = ∠ COD (Vertically opposite angles)

ΔAOB ΔCOD (SAS congruence rule)

AB = CD (By CPCT) ......... eq(1)

And,

∠OAB = ∠OCD (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel

AB || CD ............................eq(2)

From equations (1) and (2), we obtain ABCD is a parallelogram

(Theorem: Quadrilateral with any of opposite sides equal and parallel is a parallelogram.)

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠ AOD = ∠ COD (Given that each is 90⁰)

OD = OD (Common)

ΔAOD ΔCOD (SAS congruence rule)

AD = DC (3)

However,

AD = BC and

AB = CD (Opposite sides of parallelogram ABCD)

AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ΔADC and ΔBCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

ΔADC ΔBCD (SSS Congruence rule)

∠ADC = ∠BCD (By CPCT)

However,

∠ADC + ∠BCD = 180⁰ (Co-interior angles)

∠ADC + ∠ADC = 180⁰

2 ∠ADC = 180⁰

∠ADC = 90⁰ One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90⁰

Therefore, ABCD is a square

(i) ABCD is a parallelogram.

∠DAC = ∠BCA (Alternate interior angles) ... (1)

And,

∠BAC = ∠DCA (Alternate interior angles) ... (2) However, it is given that AC bisects A

∠DAC = ∠BAC ... (3)

From equations (1), (2), and (3), we obtain

∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4)

∠DCA = ∠BCA

Hence, AC bisects C

(ii) From equation (4), we obtain

∠DAC = ∠DCA

DA = DC (Side opposite to equal angles are equal)

However,

DA = BC and AB = CD (Opposite sides of a parallelogram)

AB = BC = CD = DA

Hence, ABCD is a rhombus

Let us join AC.

To Prove: ∠1 = ∠4, ∠2 = ∠3
Proof:

In ΔABC,

BC = AB (Sides of a rhombus are equal to each other)

∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)

However,

∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)

∠2 = ∠3

Therefore, AC bisects ∠C

Also,

∠2 = ∠4 (Alternate interior angles for || lines BC and DA)

∠1 = ∠4

Therefore,

AC bisects ∠A

Similarly, it can be proved that BD bisects B and D as well.

Given: ABCD is a rectangle,

∠A = ∠C

∠A = ∠C
To Prove: ABCD is a square
Proof:

∠DAC =∠ DCA (AC bisects A and C)

CD = DA (Sides opposite to equal angles are also equal)

However,

DA = BC and AB = CD (Opposite sides of a rectangle are equal)

AB = BC = CD = DA

ABCD is a rectangle and all of its sides are equal.

Hence, ABCD is a square

(ii) Let us join BD

In ΔBCD,

BC = CD (Sides of a square are equal to each other)

∠CDB = ∠CBD (Angles opposite to equal sides are equal)

However,

∠CDB = ∠ABD (Alternate interior angles for AB || CD)

∠CBD = ∠ABD

BD bisects B

Also,

∠CBD = ∠ADB (Alternate interior angles for BC || AD)

∠CDB = ∠ABD

BD bisects D

(i) In ΔAPD and ΔCQB,

∠ADP = ∠CBQ (Alternate interior angles for BC || AD)

AD = CB (Opposite sides of parallelogram ABCD)

DP = BQ (Given)

ΔAPD ΔCQB (Using SAS congruence rule)

(ii) As we had observed that,

ΔAPD ΔCQB

AP = CQ (CPCT)

(iii) In ΔAQB and ΔCPD,

∠ABQ = ∠CDP (Alternate interior angles for AB || CD)

AB = CD (Opposite sides of parallelogram ABCD)

BQ = DP (Given)

ΔAQB ΔCPD (Using SAS congruence rule)

(iv) As we had observed that,

ΔAQB ΔCPD,

AQ = CP (CPCT)

(v) From the result obtained in (ii) and (iv),

AQ = CP and AP = CQ

Since,

Opposite sides in quadrilateral APCQ are equal to each other,

APCQ is a parallelogram

(i) In ΔAPB and ΔCQD,

∠APB = ∠CQD (Each 90°)

AB = CD (Opposite sides of parallelogram ABCD)

∠ABP= ∠CDQ (Alternate interior angles for AB || CD)

ΔAPB ΔCQD (By AAS congruency)

(ii) By using the above result

ΔAPB ΔCQD, we obtain

AP = CQ (By CPCT)

(i) Given that: AB = DE and

AB || DE

If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.

Therefore, quadrilateral ABED is a parallelogram

(ii) Again,

BC = EF and BC || EF

Therefore, quadrilateral BCEF is a parallelogram

(iii) As we had observed that ABED and BEFC are parallelograms

Therefore,

AD = BE and AD || BE

(Opposite sides of a parallelogram are equal and parallel)

And,

BE = CF and BE || CF

(Opposite sides of a parallelogram are equal and parallel)

AD = CF and AD || CF

(iv) As we had observed that one pair of opposite sides (AD and CF) of quadrilateral

ACFD are equal and parallel to each other, therefore, it is a parallelogram

(v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other

AC || DF and AC = DF

(vi) ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (ACFD is a parallelogram)

ΔABC ΔDEF (By SSS congruence rule)

Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E

AD||CE ( by construction)
AE||DC ( as AB is extended to E)
It is clear that AECD is a parallelogram

(i) AD = CE (Opposite sides of parallelogram AECD)

However,

AD = BC (Given)

Therefore,

BC = CE

∠CEB = ∠CBE (Angle opposite to equal sides are also equal)

Consider parallel lines AD and CE. AE is the transversal line for them

∠A + ∠CEB = 180⁰ (Angles on the same side of transversal)

∠A + ∠CBE = 180⁰ (Using the relation CEB = CBE) ...(1)

However,

∠B + ∠CBE = 180⁰ (Linear pair angles) ...(2)

From equations (1) and (2), we obtain

∠A = ∠B

(ii) AB || CD

∠A + ∠D = 180⁰ (Angles on the same side of the transversal)

Also,

∠C + ∠B = 180⁰ (Angles on the same side of the transversal)

∠A + ∠D = ∠C + ∠B

However,

∠A = ∠B [Using the result obtained in (i)

∠C = ∠D

(iii) In ΔABC and ΔBAD,

AB = BA (Common side)

BC = AD (Given)

∠B = ∠A (Proved before)

ΔABC ΔBAD (SAS congruence rule)

(iv) We had observed that,

ΔABC ΔBAD

AC = BD (By CPCT)

(i) In ΔADC, S and R are the mid-points of sides AD and CD respectively

Mid point theorem: In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it

SR || AC and SR = AC (1) [By mid-point theorem]

(ii) In ΔABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem

PQ || AC and PQ = AC (2)

Using equations (1) and (2), we obtain

PQ || SR and PQ = SR (3)

PQ = SR

(iii) From equation (3), we obtained

PQ || SR and

PQ = SR

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal

Hence, PQRS is a parallelogram

The figure is given below:

To Prove: PQRS is a rectangle.
For that, we need to prove that, PQ = RS, PS = QR and
also, that angles of PQRS are all right angles.
Proof:
We start by proving the equality of sides of quadrilateral PQRS

In ΔABC, P and Q are the mid-points of sides AB and BC respectively,
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints
of those sides is parallel to the third side and is half the length of the third side.

∴ From the mid-point theorem,

PQ || AC &
PQ = ½ AC ..(1)

Also, R and S are the mid-points of CD and AD respectively

∴ From the mid-point theorem,

RS || AC
& RS = ½ AC ..(2)

Therefore,

From equations (1) and (2), we obtain

PQ || RS and PQ = RS

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other
So it is a parallelogram.

Let the diagonals of rhombus ABCD intersect each other at point O

In quadrilateral OMQN,

MQ || ON (Because PQ || AC)

QN|| OM (Because QR || BD)

Therefore,

OMQN is a parallelogram

∠MQN = ∠NOM

∠PQR = ∠NOM

However, NOM = 90⁰ (Diagonals of a rhombus are perpendicular to each other)

So, PQR = 90⁰

Clearly, PQRS is a parallelogram having one of its interior angles as 90⁰

Hence, PQRS is a rectangle
Hence, Proved.

Let us join AC and BD.

In ΔABC,

P and Q are the mid-points of AB and BC respectively

PQ || AC and PQ = AC (Mid-point theorem) (1)

Similarly in ΔADC,

SR || AC and SR = AC (Mid-point theorem) (2)

Clearly,

PQ || SR and PQ = SR

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram

PS || QR and PS = QR (Opposite sides of parallelogram) (3)

In ΔBCD, Q and R are the mid-points of side BC and CD respectively.

QR || BD and QR = BD (Mid-point theorem) (4)

However, the diagonals of a rectangle are equal.

AC = BD (5)

By using equation (1), (2), (3), (4), and (5), we obtain

PQ = QR = SR = PS

Hence, PQRS is a rhombus

Let EF intersect DB at G

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side

In ΔABD,

EF || AB and E is the mid-point of AD

Therefore,

G will be the mid-point of DB

EF || AB and

AB || CD

EF || CD (Two lines parallel to the same line are parallel to each other)

In ΔBCD,

GF || CD and

G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC

ABCD is a parallelogram.

So, AB || CD

Then,

AE || FC

Again,

AB = CD (Opposite sides of parallelogram ABCD)

AB = CD

AE = FC (E and F are mid-points of side AB and CD)

In quadrilateral AECF, one pair of opposite sides is parallel and equal to each other

Therefore, AECF is a parallelogram

AF || EC (Opposite sides of a parallelogram)

In ΔDQC,

F is the mid-point of side DC and FP || CQ

Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ

DP = PQ (1)

AB = CD

Similarly,

In ΔAPB,

E is the mid-point of side AB and EQ || AP

Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB

PQ = QB (2)

From equations (1) and (2),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD

Mid point theorem:The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.
Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD

In ΔABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem:

SP || BD and SP = BD (1)

Similarly in ΔBCD,

QR || BD and QR = BD (2)

From equations (1) and (2), we obtain

SP || QR and SP = QR

In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other

Therefore, SPQR is a parallelogram.

We know that diagonals of a parallelogram bisect each other

Hence, PR and QS bisect each other.

(i) In ,

Given that: M is the mid-point of AB and

MD || BC

Therefore,

D is the mid-point of AC (By converse of mid-point theorem)

(ii) As DM || CB and AC is a transversal line for them

Therefore,

∠MDC + ∠DCB = 180⁰ (Co-interior angles)

∠MDC + 90⁰ = 180⁰

∠MDC = 90⁰

MD is perpendicular to AC

(iii) Join MC

In ΔAMD and ΔCMD,

AD = CD (D is the mid-point of side AC)

∠ADM = ∠CDM (Each 90⁰)

DM = DM (Common)

ΔAMD ΔCMD (By SAS congruence rule)

Therefore,

AM = CM (By CPCT)

However,

AM = AB (M is the mid-point of AB)

Therefore,

CM = AM = AB