Quadrilaterals: class 9 math: 9 Math

mathematics Class Nine

Theorem 8.8 to 8.12: 9th math

Theorem 8.8 (Class nine math Quadrilateral)

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Solution

As given, Let PQRS is a quadrilateral and PR and SQ are its diagonals.

And, OP = OR

And OS = OQ

Then to prove that PQRS is a parallelogram.

i.e. to prove PQ||SR and PS||QR

class 9th math quadrilaterals theorem 8.8

Proof

In between ΔPOQ and ΔSOR,

OS = OQ

And OP = OR

[Because diagonals bisect each other]

And ∠POQ = ∠SOR

[Because ∠POQ and ∠SOR are vertically opposite angles and hence are equal]

Thus, from SAS (Side Angle Side) congruency,

ΔPOQ ≅ ΔSOR

Now, by CPCT, corresponding parts of congruent angles are equal

Thus, ∠OSR = ∠OQP - - - - - (i)

Now transversal SQ intersecting two lines PQ and SR in such a maaner that pairs of alternate angles are equal.

Thus, by one of the theorem of parallel lines which says that if a transversal intersects two lines in such a way that a pair of alternate internal angles are equal, then lines are parallel.

Thus, PQ||SR

Similarly, in between ΔPOS and ΔROQ,

OP = OR and OQ = OS

[As diagonals bisect each other]

And ∠POS and ∠ROQ are vertically opposite angles and are equal

i.e. ∠POS = ∠ROQ

Thus, from SAS (Side Angle Side) Congruency,

&Delta POS ≅ ΔROQ

Now, by CPCT we know that corresponding parts of congruent triangles are equal

Thus, ∠OSP = ∠OQR - - - - (ii)

Here, a transversal SQ is intersecting two lines PS and QR forming equal alternate interior angles

Thus, PS||QR

Now, since PQ||SR and PS||QR

Thus, PQRS is a parallelogram Proved

Thus, If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Theorem 8.9 (Class nine math Quadrilateral)

Each angle of a rectangle is a right angle.

Solution

Let as given, PQRS is a rectangle

class 9th math quadrilaterals theorem 8.9

Then to prove that

∠P = ∠Q = ∠R = ∠S = 90^o

Proof

Since PQRS is a rectangle

Thus, PQ||SR and PS || QR

And one of the angle is right angle.

Let ∠P = 90^o

[Because we know that a rectangle is a parallelogram in which one angle is a right angle]

Now, we know that opposite angles of a parallelogram are equal.

Thus, ∠R = ∠P

⇒ ∠R = 90^o

Now we know from one of the theorems of parallel lines which says that if a transversal bisects two parallel lines, the each pair of interior angles of the same side of the transversal is supplementary.

Here, since PS || QR

And a transversal PR is intersecting them

Therefore, ∠P + ∠Q = 180^o

[Interior angles on the same side of transversal]

⇒ 90^o + ∠Q = 180^o

⇒ ∠Q = 180^o – 90^o

⇒ ∠Q = 90^o

And, since PQRS is a parallelogram, then opposite angles Q and S are equal

i.e. ∠S = ∠Q

⇒ ∠S = 900^o

Therefore, ∠P = ∠Q = ∠R = ∠S = 90^o Proved

Thus, Each angle of a rectangle is a right angle. Proved

Theorem 8.10 (Class nine math Quadrilateral)

Diagonals of a rhombus are perpendicular to each other.

Solution

As given, let PQRS is a rhombus and PR and SQ are its diagonals.

Then to prove diagonals PR and SQ are perpendicular to each other.

That is ∠POQ = 90^o

class 9th math quadrilaterals theorem 8.10

Proof

We know that all sides are equal in a rhombus.

Thus, in rhombus PQRS,

PQ = QR = RS = PS

Again we know that a rhombus is a parallelogram, and diagonals of a parallelogram bisects each other.

Thus, in the given rhombus

OP = OR

Now, in ΔPOQ and ΔQOR,

OP = OR

And PQ = QR [Sides of rhombus]

And OQ is common side

Thus, from SSS (Side Side Side) congruency

ΔPOQ ≅ ΔQOR

Now from CPCT we know that corresponding parts of congruent triangles are equal.

Thus, ∠POQ = ∠QOR

And, since ∠POQ and ∠QOR together form linear pair of angles

Thus, ∠POQ + ∠QOR = 180^o

⇒ ∠POQ + ∠POQ = 180^o

[Since, ∠POQ = ∠QOR]

⇒ 2 ∠POQ = 180^o

⇒ ∠POQ = 180^o/2

⇒ ∠ = POQ = 90^o

Thus, ∠QOR = 90^o

Now, we know that vertically opposite angles are equal

Thus, ∠POQ = ∠ROS = ∠QOR = ∠POS = 90^o

Thus, diagonals PR and SQ are perpendicular to each other Proved

Thus, diagonals of a rhombus are perpendicular to each other. Proved

The Mid-Point Theorem

Theorem 8.11 - The Mid-Point Theorem (Class nine math Quadrilateral)

The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.

Given, Let ABC is a triangle

And, E and F are mid-points of side AB and AC

class 9th math quadrilaterals mid-point theorem

Thus, to prove that EF||BC

Construction

A line CF parallel to AB is drawn

And, DE is extended which intersects CF at point G

class 9th math quadrilaterals mid-point theorem proof

Proof

In triangle ADE and triangle ECG

E is the mid-point of AC [As given]

Therefore, AE = EC

AB||CF [According to construction]

And, AC is a transversal intersecting parallel lines AB and CF

Therefore, alternate interior angles will be equal.

That is ∠DAE = ∠ECG

And, ∠DEA and ∠GEC are vertically opposite angles and hence are equal.

That is ∠DEA = ∠GEC

Now, by ASA congruency criterion

ΔADE ≅ ΔECG

Now, by CPCT, we know that corresponding sides of congruent triangles are equal.

Therefore, DE = EG - - - - (i)

And, AD = CG

⇒ AD = CG = DB

[Because D is the mid-point of AB and thus AD = DB]

Now, since DB = CG and DB||CG

And we know that if a pair of opposite sides is equal and parallel in a quadrilateral, then it is a parallelogram.

Therefore, DBCG is a parallelogram.

Therefore, DG||BC

Consequently, DE||BC - - - - (ii)

Now, since DBCG is a parallelogram and we know that opposite sides of a parallelogram are equal.

Therefore, DG = BC

⇒ DE + EG = BC

⇒ DE + DE = BC

[From equation (i)]

⇒ 2DE = BC

⇒ DE = `1/2`BC - - - - (iii)

Now, from equation (ii) and equation (iii)

DE||BC and DE=`1/2`BC

Thus, The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it. Proved

Theorem 8.12 - Converse of the Mid-Point Theorem 8.11 (Class nine math Quadrilateral)

The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

Given, ABC is a triangle

class 9th math quadrilaterals converse of mid-point theorem

And, D is the mid-point of side AC

A line t is drawn through the mid-point D is parallel to another side AB

That means t||AB

Thus, to prove t bisects the third side BC.

Construction

class 9th math quadrilaterals converse of mid-point theorem proof

Let a line BG parallel to AC has been drawn from B

This line BG intersects line t at point F

Proof

Since, DF||AB [as given]

And, AC||BG [As per construction]

Therefore, ABFD is a parallelogram

[Since each pair of opposite sides are parallel in quadrilateral ABFD]

Therefore, BF = AD

⇒ BF = DC - - - - (i)

[Because D is the mid-point of AC as given]

Now, in between ΔDEC and ΔEBF

AC||BG and a transversal BC is intersecting it. Therefore alternate interior angles will be equal.

Therefore ∠EBF = ∠ECD

And, AC||BG and a transversal DF is intersecting it. Therefore alternate interior angles will be equal.

Therefore, ∠CDE = ∠BFE

And, BF = DC [From equation (i)]

Now, by ASA congruency criterion

ΔDEC ≅ EBF

Now by CPCT, we know that corresponding parts of congruency triangles are equal.

Therefore, BE = EC

Thus, line t bisects side BC Proved

Thus, The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. Proved

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