In this page we have *NCERT book Solutions for Class 9th Maths:Quadrilaterals* for
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ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that

(i) SR || AC and SR =1/2 AC

(ii) PQ = SR

(iii) PQRS is a parallelogram

(i) In Δ ACD

S is the mid point of AD

R is the mid point of CD

By Mid point theorem

SR || AC and SR =1/2 AC --(X)

(ii)

In Δ ABC

P is the mid point of AB

Q is the mid point of BC

By Mid point theorem

PQ || AC and PQ =1/2 AC --(Y)

From equation X and Y

PQ = SR

and PQ || SR||AC

(iii) In Quadrilaterals PQRS

PQ=SR

and PQ||SR

Hence PQRS is a parallelogram

ABCD is a rhombus and P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

The figure is shown as below

To Prove: quadrilateral PQRS is a rectangle

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR

PQ ||SR

Hence PQSR is a parallelogram

Now in Δ BCD

Q and R are mid points of sides BC and DC

By Mid point theorem

BD || QR

or

QM || OL

Also we already proved that

PQ ||SR

or LQ ||OM

In LQOM

QM || OL

LQ ||OM

Therefore LQOM is a parallelogram

Now,we know that opposite angles in the parallelogram are equal

∠ LQM = ∠ LOM

Now ∠ LOM=90 as diagonals bisects at right angle in Rhombus

Therefore

∠ LQM=90°

or

∠ PQR=90°

Now

PQRS is a parallelogram with one angle as right angle. Hence PQRS is a rectangle

ABCD is a rectangle and P, Q, R and S are mid points of the sides AB, BC, CD and DA respectively. Show that quadrilateral PQRS is a rhombus.

The figure is shown as below

To Prove: quadrilateral PQRS is a rhombus

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR

PQ ||SR

Hence PQSR is a parallelogram

Now in Δ BCD

Q and R are mid points of sides BC and DC

By Mid point theorem

QR= 1/2 BD

Nowe AC = BC

Hence

PQ=SR=QR

Now a parallelogram whose adjacent sides are equal is a rhombus.

Hence proved

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F . Show that F is the mid-point of BC.

Let G be the point on BD where EF intersect

Now in Δ ABD

E is the mid-point of AD

EG ||AB

By converse of Mid-point theorem

G is the mid-point of BD

Now in Δ BDC

G is the mid-point of BD

FG || BC

By converse of Mid-point theorem

G is the mid-point of BC

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

AB=CD ( Opposite side of parallelogram)

Since E and F are mid-points of AB and DC respectively.

AE = 1/2 AB and CF = 1/2 DC

Therefore

AE = CF and AE || CF.

One pair of opposite sides is parallel and equal

Hence AECF is a parallelogram.

AF ||EC

In Δ BAP

E is the mid-point of AB

Now AF ||EC

So, EQ || AP

Now Converse of mid-point theorem

Q is mid-point of PB

PQ = QB ---(X)

Similarly, in ΔDQC, P is the mid-point of DQ

DP = PQ ---(Y)

From (X) and (Y), we have DP = PQ = QB

Show that the line segments joining the mid- point of the opposite sides of a quadrilateral bisect each other.

To Prove : EG and FH bisect each other.

Construction : Join EF, FG, GH, HE and AC

Proof:

In Δ ABC

E and F are mid-points of AB and BC respectively.

By Mid point theorem

EF = 1/2 AC and EF || AC ----(i)

In Δ ADC

H and G are mid-points of AD and CD respectively.

By Mid point theorem

HG = 1/2 AC and HG || AC ----(ii)

From (i) and (ii), we get

EF = HG

and EF || HG

Hence EFGH is a parallelogram

Now We know that Diagonals of the parallelogram bisects each other

Here EG and FH are diagonals of the parallelogram EFGH.

Hence Proved

ABC is a triangle right angles at C. A line through the mid -point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

- D is mid- point of AC
- MD ⊥ AC
- $CM = MA = \frac {1}{2} AB$

Construction : Join CM.

Proof:

(i) In Δ ABC

M is the mid-point of AB.

BC || DM

By converse of Mid-point theorem

D is the mid-point of AC

(ii)

Now BC || DM

Hence Corresponding angles]

∠ ADM = ∠ACB

Now ∠ACB=90°

Hence

∠ ADM =90°

Hence MD ⊥ AC

(iii) In Δ ADM and CMD

AD=DC

∠ ADM = ∠ CDM = 90°

DM is common

Therefore by SAS congruence

&Delta ADM ≅ &Delta CDM

Now by CPCT

CM=MA

Since M is mid-point of AB,

MA = 1/2 AB

Hence, $CM = MA = \frac {1}{2} AB$

**Notes****Assignments & NCERT Solutions**

Class 9 Maths Class 9 Science