Chapter 8: Quadrilaterals

According to the question,

ΔABC with ∠A = 90^o and

Since, ABC is an isosceles triangle,

We get,

AB = AC …(i)

Let ADEF be the square inscribed in the isosceles triangle ABC.

Then, we have,

AD = AF = EF = AD …(ii)

Subtracting equation (ii) from (i),

AB – AD = AC – AF

BD = CF

Now,

Considering ΔCFE and ΔEDB,

BD = CF

DE = EF

∠CFE = ∠EDB = 90^o (Since, they are the side of a square)

ΔCEF ~ ΔBED (By SAS criteria)

Hence, CE = BE

Therefore, vertex E of the square bisect the hypotenuse BC.

According to the question,

We have,

ABCD is a parallelogram

AB = 10 cm

AD = 6cm.

The bisector of ∠A meets DC at E.

AE and BC produced meet at F.

Since, AF bisects ∠A,

We get,

∠BAE = ∠EAD … (1)

∠EAD = ∠EFB … (2) [Alternate angles]

From equations (1) and (2),

We get,

∠BAE = ∠EFB

Since sides opposite to equal angles are equal,

We get,

BF = AB

Here, AB = 10 cm

So, BF = 10 cm

⇒ BC + CF = 10 cm

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

⇒ CF = 10 – 6 cm = 4 cm

⇒ CF = 4 cm

According to the question,

We have,

P is the mid-point of the sides AB

Q is the mid-point of the sides BC

R is the mid-point of the sides CD

S is the mid-point of the sides DA

Also, we know that,

AC = BD.

In ΔADC, by mid-point theorem,

SR = ½ AC

And, SR||AC

In ΔABC, by mid-point theorem,

PQ = ½ AC

And, PQ||AC

Hence, SR = PQ = ½ AC

Similarly,

In ΔBCD, by mid-point theorem,

RQ = ½ BD

And, RQ||BD

In ΔBAD, by mid-point theorem,

SP = ½ BD

And, SP||BD

So, we get,

SP = RQ = ½ BD = ½ AC

Then,

SR = PQ = SP = RQ

Hence, PQRS is a rhombus.

According to the question,

We have,

P is the mid-point of the sides AB

Q is the mid-point of the sides BC

R is the mid-point of the sides CD

S is the mid-point of the sides DA

Also,

AC ⊥ BD

∠COD = ∠AOD = ∠AOB = ∠COB = 90^o

In ΔADC, by mid-point theorem,

SR = ½ AC

And, SR||AC

In ΔABC, by mid-point theorem,

PQ = ½ AC

And, PQ||AC

So, we have,

PQ||SR and SR = PQ = ½ AC

Similarly,

SP||RQ and SP = RQ = ½ BD

Now, in quadrilateral EOFR,

OE||FR and OF||ER

So, we get,

∠EOF = ∠ERF = 90^o

Hence, PQRS is a rectangle.

According to the question,

We have,

P is the mid-point of the sides AB

Q is the mid-point of the sides BC

R is the mid-point of the sides CD

S is the mid-point of the sides DA

Also,

AC ⊥ BD

And AC = BD

In ΔADC, by mid-point theorem,

SR = ½ AC

And, SR||AC

In ΔABC, by mid-point theorem,

PQ = ½ AC

And, PQ||AC

So, we have,

PQ||SR and PQ = SR = ½ AC

Now, in ΔABD, by mid-point theorem,

SP||BD and SP = ½ BD = ½ AC

In ΔBCD, by mid-point theorem,

RQ||BD and RQ = ½ BD = ½ AC

SP = RQ = ½ AC

PQ = SR = SP = RQ

Thus, we get that,

All four sides are equal.

Considering the quadrilateral EOFR,

OE||FR, OF||ER

∠EOF = ∠ERF = 90^o (Opposite angles of parallelogram)

∠QRS = 90^o

Hence, PQRS is a square.

Let the parallelogram be = ABCD

Diagonal AC bisect ∠A.

∠CAB = ∠CAD

Now,

AB||CD and AC is a transversal.

∠CAB = ∠ACD

Again, AD||BC and AC is a transversal.

∠DAC = ∠ACB

Now,

∠A = ∠C

½ ∠A = ½ ∠C

∠DAC = ∠DCA

AD = CD

But, AB = CD and AD = BC (Opposite sides of parallelograms)

AB = BC = CD = AD

Thus, ABCD is a rhombus.

According to the question,

Q is the midpoint of AB

P is the midpoint of CD

Now,

AB||CD,

Also,

AP||QC

And, AB = DC

½ AB = ½ DC

AP = QC

Now,

AP||QC and AP = QC

APCQ is a parallelogram.

AQ||PC or SQ||PR

Again,

AB||DC means ½ AB = ½ DC

BP = QD

Now, BP||QD and BP = QD

BPDQ is a parallelogram

So, PD||BQ or PS||QR

Thus, SQ||RP and PS||QR

PQRS is a parallelogram.

According to the question,

We have,

Quadrilateral ABCD

AB||CD and AD = BC.

To prove: ∠A = ∠B and ∠C = ∠D.

Construction: Draw DP ⊥ AB and CQ ⊥ AB.

Proof: In ΔAPD and ΔBQC,

Since ∠1 and ∠2 are equal to 90^o

∠1 = ∠2

Distance between parallel line,

AB = BC [Given]

By RHS criterion of congruence,

We have

ΔAPD ≅ ΔBQC [CPCT]

∠A = ∠B

Now, DC||AB

Since, sum of consecutive interior angles is 180^o

∠A+∠3 =180 …(1)

And,

∠B +∠4 =180 …(2)

From equations (1) and (2),

We get

∠A + ∠3 = ∠B + ∠4

Since, ∠A = ∠B,

We have,

⇒ ∠3 = ∠4

⇒ ∠C = ∠D

Hence, proved.

According to the question,

In quadrilateral ABED,

We have,

AB||DE and AB = DE

ABED is a parallelogram.

AD||BE and AD = BE

In quadrilateral ACFD,

We have,

AC||FD and AC = FD

ACFD is a parallelogram.

AD||CF and AD = CF

AD = BE = CF and CF||BE

In quadrilateral BCFE,

BE = CF and BE||CF.

BCFE is a parallelogram.

BC = EF and BC||EF

Hence proved.