Quadrilaterals: class 9 math: 9 Math

Theorem 8.3 to 8.7: 9th math

Theorem 8.3 (Class nine math Quadrilateral)

In a parallelogram, opposite sides are equal.

As given, let PQRS is a parallelogram

Then to prove PS = QR

And PQ = SR

Construction Points P and R has been joined.

Proof

Since PQRS is a parallelogram

Therefore, PQ||SR and PS||QR

[Because opposite sides are parallel in a parallelogram]

Now, since PS and QR are parallel lines and a transversal PR is intersecting, them.

Thus, ∠SPR and ∠PRQ are a pair of alternate interior angles.

Therefore, according to one of the theorem of parallel lines which says that if a transversal intersects two parallel lines, then each pair of alternate interior angles are equal .

Thus, ∠SPR = ∠PRQ - - - - - (i)

Similarly, a pair of alternate interior angles RPQ and PRS are equal.

That is, ∠RPQ = ∠PRS - - - - (ii)

[Pair of alternate angles]

Now, in ΔPQR and ΔPRS,

∠SPR = ∠PRQ [From equation (i)]

And, ∠RPQ = ∠PRS [From equation (ii)]

And, side PR is common in both of the triangles

Therefore, by ASA (Angle Side Angle) congruency

ΔPQR ≅ &DeltaPRS

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

Thus, PS = QR and PQ = SR Proved

Thus, In a parallelogram, opposite sides are equal. Proved

Theorem 8.4 (Class nine math Quadrilateral)

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

As given, Let PQRS is a quadrilateral.

And, its opposite sides PS = QR and PQ = SR

Therefore, to prove PQRS is a parallelogram.

Construction

A diagonal PR has been drawn

Proof

In ΔPRS and ΔPQR

PS = QR [As given in theorem]

And, PQ = SR [As given in theorem]

And, PR is a common side in both of the triangles.

Therefore, By SSS (Side Side Side) Congruency

ΔPRS ≅ ΔPQR

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

Thus, ∠1 = ∠4

And, ∠3 = ∠2

Now, we know from one of the theorem of parallel lines which says that If a transversal intersects two lines in such a way that each pair of alternate interior angles are equal, then lines are parallel.

Here, transversal PR intersects given two lines PS and QR in such a way that each pair of alternate internal angles i.e. ∠1 and 4 and angle 3 and 2 are equal.

Thus, PS||QR and PQ||SR

Since, opposite sides are parallel in the given quadrilateral, therefore PQRS is a parallelogram. Proved

Thus, A quadrilateral is a parallelogram if its opposite sides are equal Proved

Theorem 8.5 (Class nine math Quadrilateral)

In a parallelogram, opposite angles are equal.

As given, Let PQRS is a parallelogram

Thus, SR || PQ and PS ||QR

[Since opposite sides of a parallelogram are parallel.]

Thus, to prove that

∠PQR = ∠RSP

And, ∠SPQ = ∠QRS

Construction

Points P and R and points S and Q has bee joined.

Proof

Now, in ΔPRS and ΔPQR

PS||QR

And PRs is a transversal which intersects these two parallel sides or line at point P and R respectively.

Now, we know that "if a transversal intersects two parallel lines, then each pair of alternate interior angles are equal."

Here, ∠SPR and ∠QRP are a pair of alternate interior angles, hence are equal.

i.e. ∠SPR = ∠QRP

Similarly, ∠RPQ and ∠PRS are a pair of alternate interior angles

Thus, ∠RPQ = ∠PRS

And, side PR is common in both of the triangles

Thus, here two angles and side between them are equal in both of the triangles.

Thus, according to Angle Side Angle (ASA) congruency

ΔPRS ≅ ΔPQR

Thus, by CPCT (Corresponding parts of Congruent Triangles),

∠PQR = ∠RSP - - - - - (i)

Similarly, in between ΔPQS and ΔQRS

SR||PQ

And a transversal SQ is intersecting these parallel lines SR and PQ.

Thus, ∠PQS = ∠QSR

And, ∠QSP = ∠SQR

[These are pair of alternate interior angles formed by a transversal SQ intersecting two parallel lines SR and PQ]

And, side SQ is common both of the triangles PQS and ΔQRS

Thus, by ASA (Angle Side Angle) congruency,

ΔPQS ≅ ΔQRS

Thus, by CPCT (Corresponding Parts of Congruent Triangles),

∠SPQ = ∠QRS - - - - - (ii)

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

Thus, ∠PQR = ∠RSP and ∠SPQ = ∠QRS Proved

Thus, In a parallelogram, opposite angles are equal. Proved

Theorem 8.6 [Converse of theorem 8.5] (Class nine math Quadrilateral)

In a quadrilateral, if each pair of opposite angles is equal, then it is a parallelogram.

Solution

As given, Let PQRS is a quadrilateral in which opposite angles are equal

i.e. ∠P = ∠R and ∠Q = ∠S

Therefore, to prove PQRS is a parallelogram.

Proof

Here, as given, ∠P = ∠R - - - - - (i)

And, ∠Q = ∠S - - - - - - (ii)

Therefore, ∠P + ∠Q = ∠R + ∠S - - - - -(iii)

Now, we know that sum of all the four internal angles of a quadrilateral is equal to 360^o

Therefore, ∠P + ∠Q + ∠ R + ∠S = 360^o

⇒ (∠P + ∠Q) + (∠R ∠S) = 360^o

Now from equation (iii), we get

⇒ (∠P + ∠Q) + (∠P + ∠Q) = 360^o

⇒ 2 (∠P + ∠Q) = 360^o

⇒ ∠P + ∠Q = 360^o/2

⇒ ∠P + ∠Q = 180^o

⇒ ∠P + ∠Q = ∠R + ∠S = 180^o - - - - (iv)

[Because from equation (iii) we have ∠P + ∠Q = ∠R + ∠S]

Now, we know from one of theorems of parallel lines which says that If a transversal intersects two lines such that a pair of interior angles on the same side of transversal is supplementary, then the two lines are parallel.

Here, a transversal PQ is intersecting two lines PS and QR in such a way that a pair of interior angles P and Q on the same side of transversal is supplementary (from equation (iv), thus lines PS are QR are parallel.

That means PS||QR

Now, since ∠P + ∠Q = 180^o [from equation (iv)]

⇒ ∠R + ∠Q = 180^o

[Because as given that opposite angles are equal, i.e. ∠P = ∠Q]

Here, a transversal QR intersecting two lines PQ and SR in such a way that a pair of interior angles P and Q on the same side of transversal is supplementary (from equation (iv), thus lines PQ are SR are parallel.

That means PQ||SR

Now, since opposite sides are parallel in the given quadrilateral, thus it is a parallelogram.

That is In a quadrilateral, if each pair of opposite angles is equal, then it is a parallelogram. Proved

Theorem 8.7 (Class nine math Quadrilateral)

The diagonals of a parallelogram bisect each other.

Solution

As given, let PQRS is a parallelogram.

And, PR and SQ are its two diagonals.

Thus, to prove that PR and SQ bisects each other.

That is OP = OR and OQ = OS

Proof

Since, PQRS is a parallelogram,

Thus, PS = QR and PS||QR

Now, In between ΔPOS and ΔROQ,

PS||QR and a transversal PR is intersecting them, thus each pair of alternate interior angles are equal.

Thus, ∠2 = ∠4 - - - - - (i)

Similarly, PS||QR and a transversal QS is intersecting them, thus each pair of alternate angles are equal.

Thus, ∠1 = ∠3 - - - - (ii)

And, PS = QR [Because opposite sides of parallelogram]

Thus, from ASA (Angle Side Angle) congruency,

ΔPOS ≅ ΔROQ

Now, form CPCT, corresponding parts of congruent triangles are equal,

Thus, OP = OR and OS = OQ Proved

Thus, The diagonals of a parallelogram bisects each other. Proved

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