Mid-point Theorem

(i) Since F is midpoint of AB and E is midpoint of AC,

∴ FE is parallel to BC and FE = $\frac{1}{2}$BC = $\frac{1}{2}(4.8)$ = 2.4 cm (By midpoint theorem)

FB = $\frac{1}{2}$AB = $\frac{1}{2}(6)$ = 3 cm

EC = $\frac{1}{2}$AC = $\frac{1}{2}(5.6)$ = 2.8 cm.

Perimeter of trapezium FBCE = FE + EC + BC + FB = 2.4 + 2.8 + 4.8 + 3 = 13 cm.

Hence, perimeter of trapezium FBCE = 13 cm.

(ii) Since F is midpoint of AB and E is midpoint of AC,

∴ FE is parallel to BC and FE = $\frac{1}{2}$BC = $\frac{1}{2}(4.8)$ = 2.4 cm (By midpoint theorem)

Since F is midpoint of AB and D is midpoint of BC,

∴ FD is parallel to AC and FD = $\frac{1}{2}$AC = $\frac{1}{2}(5.6)$ = 2.8 cm (By midpoint theorem)

Since E is midpoint of AC and D is midpoint of BC,

∴ ED is parallel to AB and ED = $\frac{1}{2}$AB = $\frac{1}{2}(6)$ = 3 cm (By midpoint theorem)

Perimeter of △DEF = FE + FD + ED = 2.4 + 2.8 + 3 = 8.2 cm

Hence, perimeter of △DEF = 8.2 cm.

(i) Since, D and E are mid-points of the sides AB and AC respectively,

∴ DE is parallel to BC and DE = $\frac{1}{2}BC = \frac{1}{2}(5.6)$ = 2.8 cm. (By midpoint theorem)

Hence, DE = 2.8 cm.

(ii) Since, DE is parallel to BC.

∴ ∠ABC = ∠ADE (Corresponding angles)

⇒ ∠ADE = 72°.

Hence, ∠ADE = 72°.

In △ABC,

D is the midpoint of AB and DF || BC

∴ F is the midpoint of AC (By converse of mid-point theorem)

F and E are midpoints of AC and BC respectively

∴ EF || AB ⇒ EF || DB .....(1)

From figure,

⇒ DF || BE ......(2)

Using 1 and 2,

⇒ EF || DB and DF || BE

Hence, proved that DBEF is a parallelogram.

F is the midpoint of AC we get,

AC = 2 × AF = 2 × 2.6 = 5.2 cm

Hence, AC = 5.2 cm

From figure,

In △ABC,

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

Now join DE, EF and FD.

To prove :

△ADF ≅ △DBE ≅ △ECF ≅ △DEF

In △ABC,

D and E are midpoints of AB and BC

∴ DE || AC or,

DE || FC .......(i)

or DE || AF .........(ii)

D and F are midpoints of AB and AC

∴ DF || BC or,

DF || EC .......(iii)

or DF || BE ........(iv)

F and E are midpoints of AC and BC

∴ FE || AB or,

FE || AD .......(v)

or FE || DB (vi)

From (i) and (iii) we get,

DE || FC and DF || EC.

∴ DECF is a parallelogram.

We know that,

Diagonal FE divides the parallelogram DECF in two congruent triangles DEF and CEF.

∴ △DEF ≅ △ECF .......(1)

From (ii) and (v) we get,

DE || AF and FE || AD.

∴ ADEF is a parallelogram.

We know that,

Diagonal FD divides the parallelogram in two congruent triangles DEF and AFD.

∴ △DEF ≅ △AFD .......(2)

From (iv) and (vi) we get,

DF || BE and FE || DB.

∴ DBEF is a parallelogram.

We know that,

Diagonal DE divides the parallelogram in two congruent triangles DEF and DBE.

∴ △DEF ≅ △DBE .......(3)

Using equations 1, 2 and 3 we get,

△ADF ≅ △DBE ≅ △ECF ≅ △DEF.

Hence, proved that four triangles formed by joining in pairs, the mid-points of the sides of a triangle are congruent to each other.

It is given that,

ABC is an isosceles triangle. Let AB = AC = x.

D, E and F are mid-points of the sides AB, BC and CA respectively.

Join D, E and F.

D and E are midpoints of AB and BC

∴ DE || AC and DE = $\frac{1}{2}$AC = $\frac{x}{2}$. (By midpoint theorem) ......(i)

F and E are midpoints of AC and BC

∴ FE || AB and FE = $\frac{1}{2}$AB = $\frac{x}{2}$. (By midpoint theorem) ......(ii)

From (i) and (ii) we get, DE = FE.

Hence, proved that △DEF is an isosceles triangle.

(i) Given,

ABCD is a parallelogram in which diagonals AC and BD intersect each other at O, P is the midpoint of AD.

Join OP.

In parallelogram, diagonals bisect each other,

∴ BO = OD.

Here, O is the mid-point of BD.

In △ABD,

P and O are midpoints of AD and BD respectively,

PO || AB and PO = $\frac{1}{2}$AB (By midpoint theorem) ......(i)

Hence, proved that PO || AB.

(ii) ABCD is a parallelogram.

∴ AB = CD .......(ii)

Using both (i) and (ii) we get,

PO = $\frac{1}{2}$AB = $\frac{1}{2}$CD.

Hence, proved that PO = $\frac{1}{2}$CD.

(i) In △ADC,

S and R are midpoints of AD and DC respectively,

∴ SR || AC and SR = $\frac{1}{2}$AC (By mid-point theorem) .....(i)

Hence, proved that SR || AC and SR = $\frac{1}{2}$AC (By mid-point theorem).

(ii) In △ABC,

P and Q are midpoints of AB and BC,

PQ || AC and PQ = $\frac{1}{2}$AC .......(ii)

Using (i) and (ii) we get,

PQ = SR and PQ || SR.

Hence, proved that PQ = SR.

(iii) Since, PQ = SR and PQ || SR.

Hence, proved that PQRS is a parallelogram.

Let ABCD be a square in which E, F, G and H are midpoints of AB, BC, CD and DA respectively.

Join EF, FG, GH and HE.

Join AC and BD.

In △ACD,

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

∴ GH || AC and GH = $\frac{1}{2}$AC .......(i)

In △ABC,

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

∴ EF || AC and EF = $\frac{1}{2}$AC .......(ii)

Using (i) and (ii) we get,

EF || GH and EF = GH = $\frac{1}{2}$AC ........(1)

In △ABD,

E and H are mid-points of AB and AD respectively,

∴ EH || BD and EH = $\frac{1}{2}$BD .......(iii)

In △BCD,

G and F are mid-points of CD and BC respectively,

∴ FG || BD and FG = $\frac{1}{2}$BD .......(iv)

Using (iii) and (iv) we get,

EH || FG and EH = FG = $\frac{1}{2}$BD ........(2)

We know that diagonals of square are equal,

AC = BD

Dividing both sides by 2 we get,

$\frac{\text{AC}}{2} = \frac{\text{BD}}{2}$

Substituting above value in 1 and 2 we get,

EF = GH = EH = FG .........(v)

∴ EFGH is a parallelogram.

In △GOH and △GOF,

OH = OF as diagonals of parallelogram bisect each other.

OG = OG (Common)

GH = GF (From (v))

∴ △GOH ≅ △GOF (SSS axiom of congruency)

∠GOH = ∠GOF (c.p.c.t.c.)

From figure,

⇒ ∠GOH + ∠GOF = 180°

⇒ ∠GOH + ∠GOH = 180°

⇒ 2∠GOH = 180°

⇒ ∠GOH = 90°.

So, the diagonals of EFGH bisect and are perpendicular to each other.

∴ EFGH is a square.

Hence, proved that quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.

In △BCE,

D is the midpoint of BC (As AD is median)

DF || BE

∴ F is the midpoint of CE (By converse of mid-point theorem).

⇒ CF = $\frac{1}{2}CE$ .......(i)

Given,

BE is median

∴ CE = $\frac{1}{2}AC$

Substituting value of CE in (i) we get,

$$\begin{aligned}\Rightarrow CF = \frac{1}{2}CE \\[1em] \Rightarrow CF = \frac{1}{2} \times \frac{1}{2}AC \\[1em] \Rightarrow CF = \frac{1}{4}AC.\end{aligned}$$

Hence, proved that $CF = \frac{1}{4}AC.$

(i) We know that,

ABCD is a parallelogram,

∴ FC || BE

∠CEB = ∠FCE (Alternate angles)

⇒ ∠HEB = ∠FCH .......(1)

∠EBF = ∠CFB (Alternate angles)

⇒ ∠EBH = ∠CFH .......(2)

Here E and F are mid-points of AB and CD

BE = $\frac{1}{2}$AB ........(3)

CF = $\frac{1}{2}$CD ........(4)

We know that ABCD is a parallelogram,

AB = CD

Now dividing both sides by $\frac{1}{2}$

$\frac{1}{2}$AB = $\frac{1}{2}$CD

Using equations 3 and 4 we get,

BE = CF .......(5)

In △HEB and △HCF,

∠HEB = ∠FCH (Using eqn. i)

∠EBH = ∠CFH (Using eqn. ii)

BE = CF (Using eqn. v)

∴ △HEB ≅ △HCF (By ASA axiom of congruency)

Hence, proved that △HEB ≅ △HCF.

(ii) AB = CD (As ABCD is a parallelogram)

Hence, AE = CF (As E and F are mid-points of the sides AB and CD respectively)

As, AB || CF we can say that,

AE || CF

Since, AE = CF and AE || CF

∴ AECF is a parallelogram.

∴ AF || EC

From figure we get,

GF || EH ........(1)

AB = CD (As ABCD is a parallelogram)

Hence, DF = EB (As E and F are mid-points of the sides AB and CD respectively) and DF || EB.

Since, DF = EB and DF || EB

∴ DEBF is a parallelogram.

∴ DE || FB

From figure we get,

GE || FH ........(2)

From 1 and 2 we get,

GF || EH and GE || FH.

∴ GEHF is a parallelogram.

Hence, proved that GEHF is a parallelogram.

From figure,

In △ABD and △ACD,

△ABC is an isosceles triangle

∴ ∠ABD = ∠ACD

Here D is the mid-point of BC

BD = CD

It is given that AB = AC

∴ △ABD ≅ △ACD (By SAS axiom of congruency)

⇒ ∠ADB = ∠ADC (By c.p.c.t.c)

From figure,

⇒ ∠ADB + ∠ADC = 180°
⇒ ∠ADB + ∠ADB = 180°
⇒ 2∠ADB = 180°
⇒ ∠ADB = 90°.

So, AD is perpendicular to BC.

D and E are mid-points of BC and AB,

By midpoint theorem,

DE || AC or,

DE || AF .......(i)

D and F are mid-points of BC and AC,

By midpoint theorem,

DF || AB or,

DF || AE .......(ii)

Using (i) and (ii) we get,

AEDF is a parallelogram.

Diagonals of parallelogram bisect each other

AD and EF bisect each other.

Since, E and F are mid-points of AB and AC,

By midpoint theorem,

EF || BC

Since, AD is perpendicular to BC and EF || BC.

∴ AD ⊥ EF.

Hence, proved that line segment AD is perpendicular to EF and is bisected by it.

(i) In △ABD,

E is mid-point of AD and F is mid-point of BD,

∴ EF || AB and EF = $\frac{1}{2}$AB .......(1)

Given,

AB || CD

Since, EF || AB and AB || CD

⇒ EF || CD

⇒ EG || CD.

Since, EG || CD we can say,

In △BCD,

⇒ FG || CD

Given, F is midpoint of BD and FG || CD

∴ G is the midpoint of BC. (By converse of mid-point theorem)

Hence, proved that G is the midpoint of BC.

(ii) In △BCD,

F and G are midpoint of BD and BC respectively,

FG = $\frac{1}{2}$CD ..........(2)

Adding eqn. (1) from part (i) and eqn (2) we get,

EF + FG = $\frac{1}{2}$AB + $\frac{1}{2}$CD

EG = $\frac{1}{2}$(AB + CD).

Hence, proved that EG = $\frac{1}{2}$(AB + CD).

(i) Given,

EG || AB, we can say that

⇒ EF || AB

In △DAB,

E is midpoint of AD and EF || AB

∴ F is midpoint of BD (By converse of mid-point theorem).

EF = $\frac{1}{2}$AB .......(1)

Given,

EG || DC we can say that,

FG || DC

In △BCD,

F is midpoint of BD and FG || DC

∴ G is midpoint of BC (By converse of mid-point theorem).

Hence, proved that G is midpoint of BC.

(ii) In △BCD,

F is midpoint of BD and G is midpoint of BC

∴ FG = $\frac{1}{2}$DC .......(2)

Adding eqn. 1 from part (i) and eqn. 2 we get,

EF + FG = $\frac{1}{2}$AB + $\frac{1}{2}$DC

EG = $\frac{1}{2}$(AB + CD)

2EG = AB + CD.

Hence, proved that 2EG = AB + CD.

ABCD is a trapezium in which AB || DC and E, F are mid-points of AD and BC respectively.

Join CE and produce it to meet BA produced at G.

In △EDC and △EAG,

ED = EA (∵ E is mid-point of AD)

∠CED = ∠ GEA (Vertically opposite ∠s)

∠ECD = ∠EGA (Alternate ∠s)

∴ △EDC ≅ △EAG

⇒ CD = GA and EC = EG (c.p.c.t.)

In △CGB,

E is mid-point of CG

F is mid-point of BC

∴ By mid-point theorem, EF || AB and EF = $\frac{1}{2}$GB.

But GB = GA + AB = CD + AB

∴ EF = $\frac{1}{2}$(AB + CD) .....(1)

(i) Given,

AB = 6 cm and DC = 4 cm,

Putting these values in eq (1) we get,

EF = $\frac{1}{2}$(6 + 4)

= $\frac{1}{2}$ x 10

= 5 cm

Hence, EF = 5 cm.

(ii) Given,

DC = 8 cm and EF = 9 cm

Putting these values in eq (1) we get,

9 = $\frac{1}{2}$(AB + 8)

⇒ 18 = AB + 8

⇒ AB = 18 - 8 = 10 cm

Hence, AB = 10 cm.

It is given that,

In △ABD,

P and Q are mid-points of AB and BD,

PQ || AD and PQ = $\frac{1}{2}$AD = $\frac{1}{2}$BC .......(i)

In △ACD,

R and S are mid-points of CD and AC,

RS || AD and RS = $\frac{1}{2}$AD = $\frac{1}{2}$BC .......(ii)

In △BCD,

R and Q are mid-points of CD and BD,

RQ || BC and RQ = $\frac{1}{2}$BC .......(iii)

In △ABC,

P and S are mid-points of AB and AC,

PS || BC and PS = $\frac{1}{2}$BC .......(iv)

From (i) and (ii) we get,

PQ || RS

From (iii) and (iv) we get,

RQ || PS

From (i), (ii), (iii) and (iv) we get,

PQ = RS = PS = RQ.

Since, all sides are equal and opposite sides are parallel.

Hence, proved that PQRS is a rhombus.

Construction,

1. Join AC and BD

2. AC and BD intersect at O.

3. Join EF and FG.

(i) We know that,

Diagonals of a kite intersect at right angles.

∠MON = 90° .......(i)

In △BCD,

E and F are mid-points of CD and BC,

EF || DB and EF = $\frac{1}{2}$DB ......(ii)

Since, EF || DB we can say that,

MF || ON.

As sum of opposite angles of a quadrilateral = 180°

∠MON + ∠MFN = 180°

90° + ∠MFN = 180°

∠MFN = 90°.

From figure,

∠EFG = ∠MFN = 90°.

Hence, proved that ∠EFG = 90°.

(ii) From part (i) we get,

FE || BD

Here line through G (GH) is parallel to FE.

∴ GH || FE

or GH || BD.

In △ABD,

GH || BD and G is midpoint of AB,

∴ H is mid-point of AD (By converse of mid-point theorem).

Hence, proved that the line drawn through G and parallel to FE bisects DA.

(i) In △ACD,

G is mid-point of CD and BG is parallel to AD,

∴ B is mid-point of AC (By converse of mid-point theorem).

By mid-point theorem,

BG = $\frac{1}{2}$AD = $\frac{1}{2}$ x 6 = 3 cm.

Hence, BG = 3 cm.

(ii) In △CDF,

G is mid-point of CD and GE || CF

∴ E is mid-point of FD (By converse of mid-point theorem).

By mid-point theorem,

GE = $\frac{1}{2}$CF

CF = 2GE

CF = 2(2.3) = 4.6 cm

Hence, CF = 4.6 cm.

(iii) From part (i)

B is mid-point of AC,

∴ AB = BC

Hence, AB = 2.4 cm

(iv) From part (ii),

E is mid-point of FD,

∴ ED = $\frac{1}{2}$FD = $\frac{1}{2}$ x 4.4 = 2.2 cm

Hence, ED = 2.2 cm

Since, D is midpoint of AB and E is midpoint of BC,

DE || AC and DE = $\frac{1}{2}$AC = $\frac{1}{2}$ x 5 = 2.5 cm

Hence, Option 3 is the correct option.

Length of diagonal (BD) of ABCD = $\sqrt{AB^2 + AD^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}$ = 10 cm.

In △BCD,

P is midpoint of BC and Q is midpoint of CD.

PQ || BD and PQ = $\frac{1}{2}$BD = $\frac{1}{2}$ x 10 = 5 cm.

Hence, Option 2 is the correct option.

In △ABC,

D and E are respective midpoints of AB and AC.

∴ DE || (BC or PQ) ......(1)

Again in △ABO,

D and P are respective midpoints of AB and BO.

∴ DP || AO ........(2)

In △ACO,

E and Q are respective midpoints of AC and CO.

∴ EQ || AO ........(3)

From 2 and 3 we get,

DP || EQ .......(4)

From 1 and 4 we get,

DE || PQ and DP || EQ

∴ DEQP is a parallelogram.

Hence, Option 4 is the correct option.

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let PR ⊥ QS.

In △QRP,

A and B are midpoints of PQ and QR respectively.

∴ AB || PR and AB = $\frac{1}{2}PR$ (By midpoint theorem) ........(1)

Similarly in △PRS,

D and C are midpoints of PS and RS respectively.

∴ DC || PR and DC = $\frac{1}{2}PR$ (By midpoint theorem) ........(2)

In △PQS,

D and A are midpoints of PS and PQ respectively.

∴ DA || QS and DA = $\frac{1}{2}QS$ (By midpoint theorem).........(3)

Similarly in △QRS,

B and C are midpoints of QR and SR respectively.

∴ BC || QS and BC = $\frac{1}{2}QS$ (By midpoint theorem).........(4)

From 1 and 2 we get,

AB = DC and AB || DC

From 3 and 4 we get,

DA = BC and DA || BC

Hence, proved that ABCD is a parallelogram.

Since, DA || QS and PR ⊥ QS

∴ DA ⊥ PR.

Since, DC || PR and PR ⊥ QS

∴ DC ⊥ QS.

In OMDN,

∠OMD = ∠SMC (Vertically opposite angle are equal)

In quadrilateral sum of angles = 360°

∠O + ∠M + ∠N + ∠D = 360°

90° + 90° + 90° + ∠D = 360°

∠D = 360° - 270° = 90°.

In parallelogram sum of alternate angles = 180°.

∠D + ∠B = 180°

90° + ∠B = 180°

∠B = 90°

Since, opposite sides are equal and adjacent sides are perpendicular to each other.

Hence, ABCD is a rectangle.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rectangle if the diagonals of PQRS are perpendicular to each other.

Hence, Option 3 is the correct option.

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let diagonals be of equal length i.e. AC = BD = x

In △BCA,

P and Q are midpoints of AB and BC respectively.

∴ PQ || AC and PQ = $\frac{1}{2}$AC = $\frac{1}{2}$x (By midpoint theorem) ........(1)

Similarly in △ACD,

S and R are midpoints of AD and CD respectively.

∴ SR || AC and SR = $\frac{1}{2}$AC = $\frac{1}{2}$x (By midpoint theorem) ........(2)

In △ABD,

S and P are midpoints of AD and AB respectively.

∴ SP || BD and SP = $\frac{1}{2}$BD = $\frac{1}{2}$x (By midpoint theorem).........(3)

Similarly in △BCD,

Q and R are midpoints of BC and CD respectively.

∴ QR || BD and QR = $\frac{1}{2}$BD = $\frac{1}{2}$x (By midpoint theorem).........(4)

From 1, 2, 3 and 4 we get,

PQ = SR = SP = QR.

Hence, proved that PQRS is a rhombus.

∴ The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a rhombus if the diagonals of ABCD are equal.

Hence, Option 3 is the correct option.

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let diagonals be of equal length i.e. AC = BD = x and AC ⊥ BD.

In △BCA,

P and Q are midpoints of AB and BC respectively.

∴ PQ || AC and PQ = $\frac{1}{2}AC = \frac{1}{2}x$ (By midpoint theorem) ........(1)

Similarly in △ACD,

S and R are midpoints of AD and CD respectively.

∴ SR || AC and SR = $\frac{1}{2}AC = \frac{1}{2}x$ (By midpoint theorem) ........(2)

In △ABD,

S and P are midpoints of AD and AB respectively.

∴ SP || BD and SP = $\frac{1}{2}BD = \frac{1}{2}x$ (By midpoint theorem).........(3)

Similarly in △BCD,

Q and R are midpoints of BC and CD respectively.

∴ QR || BD and QR = $\frac{1}{2}BD = \frac{1}{2}x$ (By midpoint theorem).........(4)

From 1, 2, 3 and 4 we get,

PQ = SR = SP = QR.

Hence, proved that PQRS is a rhombus.

Since, SP || BD and AC ⊥ BD

∴ SP ⊥ AC.

Since, SR || AC and AC ⊥ BD

∴ SR ⊥ BD.

In OMSN,

∠OMS = ∠DMR (Vertically opposite angle are equal)

In quadrilateral sum of angles = 360°

∠O + ∠M + ∠N + ∠S = 360°

90° + 90° + 90° + ∠S = 360°

∠S = 360° - 270° = 90°.

Since, in rhombus adjacent angles sum = 180°

∠S + ∠R = 180°

90° + ∠R = 180°

∠R = 90°.

∠Q + ∠R = 180°

90° + ∠Q = 180°

∠Q = 90°.

∠S + ∠P = 180°

90° + ∠P = 180°

∠P = 90°.

Since, PQ = QR = RS = SP and ∠P = ∠Q = ∠R = ∠S = 90°

Hence, proved that PQRS is a square.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if diagonals of ABCD are equal and perpendicular to each other.

Hence, Option 4 is the correct option.

According to statement 1 :

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

This statement describes mid-point theorem.

∴ Statement 1 is true.

According to statement 2 :

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

This statement describes the converse of mid-point theorem.

∴ Statement 2 is true.

Hence, option 1 is the correct option.

According to the Midpoint Theorem:

"The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of its length."

∴ Reason (R) is true.

In △ABC:

⇒ E is the midpoint of AB.

⇒ F is the midpoint of AC.

The line segment joining E and F is EF.

Therefore, by the Midpoint Theorem, EF ∥ BC.

The quadrilateral EFCB has one pair of parallel sides (EF and BC).

Therefore, the quadrilateral EFCB is indeed a trapezium.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

According to converse of mid-point theorem, the straight line drawn through the mid-point of one side of a triangle parallel to another bisects the third side.

It is given that PQ is drawn parallel to DF. And, EP = 3 cm.

∴ P is midpoint of ED.

Therefore, Q will also be mid-point of EF.

According to Midpoint Theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of it.

PQ = $\frac{1}{2}$ x DF = $\frac{1}{2}$ x 6 cm

Thus, PQ = 3 cm.

∴ Assertion (A) is true.

A line segment parallel to the base does not necessarily cut the other two sides in half unless it passes through the midpoints.

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

According to equal intercept theorem, if a tranversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

∴ Reason (R) is true.

Three lines p, q, r are parallel to each other

PQ = QR = 1 cm.

⇒ PQ = QR

By equal intercept theorem.

⇒ AB = BC

From figure, AC = AB + BC

⇒ AC = AB + AB

⇒ AC = 2AB

⇒ AB = $\frac{1}{2}$ AC.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason (or explanation) for Assertion (A).

Hence, option 3 is the correct option.

Join AC and BD.

Diagonals of rhombus intersect at right angle.

∠MON = 90°

In △BCD,

Q and R are mid-points of BC and CD.

RQ || DB and RQ = $\frac{1}{2}$DB

RQ || DB ⇒ MQ || ON

From figure,

∠MON + ∠MQN = 180° (Sum of alternate angles of quadrilateral = 180°)

∠MQN = 180° - 90° = 90°

∴ PQ ⊥ QR.

Hence, proved that PQ ⊥ QR.

From figure,

ABCD is a quadrilateral in which diagonals AC and BD are perpendicular to each other. P, Q, R and S are mid-points of AB, BC, CD and DA.

In △ABC,

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

PQ || AC and PQ = $\frac{1}{2}$AC .......(1) (By mid-point theorem)

In △ADC,

S and R are mid-points of AD and DC respectively,

SR || AC and SR = $\frac{1}{2}$AC .........(2) (By mid-point theorem)

Using eqn. 1 and 2 we get,

PQ || SR and PQ = SR.

So, PQRS is a parallelogram.

In △ABD,

S and P are mid-points of AD and AB respectively,

SP || BD and SP = $\frac{1}{2}$BD .........(3) (By mid-point theorem)

Given,

AC and BD intersect at right angles,

From 3 we get,

SP || BD.

∴ SP ⊥ AC

From 2 we get,

SR || AC

∴ SP ⊥ SR i.e. ∠RSP = 90°.

∴ PQRS is a rectangle.

Hence, proved that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.

△ABC with D, E and F as mid-points of the sides BC, CA and AB is shown below:

D and E are midpoints of BC and CA respectively,

DE = $\frac{1}{2}$AB and DE || AB or DE || AF .......(1)

Since,

F is midpoint of AB,

AF = $\frac{1}{2}$AB

∴ AF = DE .......(2)

F and D are midpoints of AB and BC respectively,

FD = $\frac{1}{2}$AC and FD || AC or FD || AE .......(3)

Since,

E is midpoint of AC,

AE = $\frac{1}{2}$AC

∴ FD = AE .......(4)

From 1, 2, 3 and 4 we get,

DE || AF, AF = DE and FD || AE, FD = AE.

Hence, AEDF is a parallelogram.

∴ AD and EF bisect each other.

Hence, AD and EF bisect each other.

Since, D and E are mid-points of AB and AC respectively,

DE || BC or DE || BF and DE = $\frac{1}{2}$BC ......(By midpoint theorem) .....(i)

Given, through E, a straight line is drawn parallel to AB to meet BC at F.

F will be mid-point of BC (By converse of mid-point theorem).

Since, F is midpoint of BC,

∴ BF = $\frac{1}{2}$BC .....(ii)

From (i) and (ii) we get,

DE = BF and DE || BF.

Since, F and E are mid-points of BC and AC respectively,

FE || AB or FE || BD and FE = $\frac{1}{2}$AB ......(By midpoint theorem) .....(iii)

Since, D is midpoint of AB,

∴ BD = $\frac{1}{2}$AB .....(iv)

From (iii) and (iv) we get,

BD = FE and BD || FE.

Since, DE = BF, DE || BF and BD = FE, BD || FE

Hence, proved that BDEF is a parallelogram.

Perimeter of BDEF = BD + DE + FE + BF = BD + DE + BD + FE = 2(BD + FE).

BD = $\frac{1}{2}$AB = $\frac{1}{2}(8)$ = 4 cm.

FE = $\frac{1}{2}$BC = $\frac{1}{2}(9)$ = 4.5 cm.

Perimeter of BDEF = 2(BD + FE) = 2(4 + 4.5) = 2 × 8.5 = 17 cm.

Hence, perimeter of BDEF = 17 cm.

Given, DL || EB.

Since, DL || BE we can say that,

⇒ BE || DF

In △AFD,

E is midpoint of AD and BE is parallel to DF,

∴ B is midpoint of AF (By converse of midpoint theorem).

In BEDL,

LD || BE and BL || DE

∴ BEDL is a parallelogram.

Since, BEDL is a parallelogram opposite sides are equal.

Let LD = BE = x.

E is midpoint of AD and B is the midpoint of AF

By midpoint theorem,

BE = $\frac{1}{2}$FD

FD = 2BE = 2x.

LF = FD - LD = 2x - x = x.

Since, LF = BE = x.

Hence, proved that B is midpoint of AF and EB = LF.

In parallelogram, diagonals bisect each other.

∴ AO = OC = $\frac{1}{2}$AC ......(i)

In △BCD,

P and Q are midpoints of CD and BC,

PQ || BD (By midpoint theorem)

Since, PQ || BD

∴ QR || BO

In △BCO,

Q is midpoint of BC and QR || BO

∴ R is midpoint of OC

CR = $\frac{1}{2}$OC

Substituting value of OC from (i) in above equation,

CR = $\frac{1}{2}$OC = $\frac{1}{2} \times \frac{1}{2} \times$ AC = $\frac{1}{4}$AC.

Hence, proved that CR = $\frac{1}{4}$AC.