Mid-Point Theorem & Intercept Theorem — Exercise 9

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

(i) If AB = 6.2 cm, find DE

(ii) If DF = 3.8 cm, find AC

(iii) If perimeter of △ABC is 21 cm, find FE

Answer

By mid-point theorem,

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

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

By mid-point theorem,

⇒ DE = $\dfrac{1}{2}AB = \dfrac{1}{2}$ × 6.2 = 3.1 cm

Hence, DE = 3.1 cm.

(i) Since, F and D are the mid-points of AB and BC respectively.

⇒ DF = $\dfrac{1}{2}$ AC

⇒ AC = 2 × 3.8

⇒ AC = 7.6 cm

Hence, AC = 7.6 cm.

(iii) From above,

AC = 7.6 cm, AB = 6.2 cm

Given,

Perimeter of △ABC = 21 cm

⇒ AB + AC + BC = 21

⇒ 6.2 + 7.6 + BC = 21

⇒ BC + 13.8 = 21

⇒ BC = 21 - 13.8

⇒ BC = 7.2 cm

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

⇒ FE = $\dfrac{1}{2}$ BC

⇒ FE = $\dfrac{1}{2}$ × 7.2

⇒ FE = 3.6 cm

Hence, FE = 3.6 cm.

In the given figure, LMN is a right triangle in which ∠M = 90°, P and Q are mid-points of LM and LN respectively. If LM = 9 cm, MN = 12 cm and LN = 15 cm, find :

(i) the perimeter of trapezium MNQP

(ii) the area of trapezium MNQP

Answer

Since, P is mid-point of LM,

LP = PM

From figure,

⇒ LM = LP + PM

⇒ 9 = PM + PM

⇒ 9 = 2 PM

⇒ PM = $\dfrac{9}{2}$

⇒ PM = 4.5 cm

Since, Q is mid-point of LN,

LQ = QN

From figure,

⇒ LN = LQ + QN

⇒ 15 = QN + QN

⇒ 15 = 2 QN

⇒ QN = $\dfrac{15}{2}$

⇒ QN = 7.5 cm

By mid-point theorem,

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

Since, P and Q are the mid-points of LM and LN respectively.

⇒ PQ = $\dfrac{1}{2}MN = \dfrac{1}{2}$ × 12 = 6 cm

(i) Perimeter of trapezium MNQP = PM + PQ + QN + MN

⇒ 4.5 + 6 + 7.5 + 12

⇒ 30 cm.

Hence, perimeter of trapezium MNQP = 30 cm.

(ii) Area of trapezium MNQP = $\dfrac{1}{2}$ × (Sum of parallel sides) × height of trapezium

= $\dfrac{1}{2}$ × (PQ + MN) × PM

= $\dfrac{1}{2}$ × (6 + 12) × 4.5

= $\dfrac{1}{2}$ × 18 × 4.5

= 40.5 cm²

Hence, area of trapezium MNQP is 40.5 cm².

In the given figure, D, E, F are respectively the mid-points of the sides AB, BC and CA of △ABC. Prove that ADEF is a parallelogram.

Answer

Given,

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

AD = DB, AF = FC and BE = EC

By mid-point theorem,

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

Since, D and E are the mid-points of AB and BC respectively.

⇒ DE || AC

⇒ DE || AF .....(1)

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

⇒ FE || AB

⇒ FE || AD .....(2)

In quadrilateral ADEF,

AD // FE and DE // AF

Since, opposite sides of quadrilateral ADEF are parallel.

∴ ADEF is a parallelogram.

Hence, proved that ADEF is a parallelogram.

If D, E, F are respectively the mid-points of the sides AB, BC and CA of an equilateral triangle ABC, prove that △DEF is also an equilateral triangle.

Answer

Given,

△ABC is an equilateral triangle.

⇒ AB = BC = AC

By mid-point theorem,

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

Since, D and E are the mid-points of AB and BC respectively.

⇒ DE = $\dfrac{1}{2}$ × AC

⇒ DE = $\dfrac{1}{2}$ × AB [As AB = AC = BC] ....(1)

Since, D and F are the mid-points of AB and AC respectively.

⇒ DF = $\dfrac{1}{2}$ × BC

⇒ DF = $\dfrac{1}{2}$ × AB [As AB = AC = BC] ....(2)

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

⇒ EF = $\dfrac{1}{2}$ × AB ....(3)

From eq.(1), (2) and (3), we have:

⇒ DE = DF = EF

∴ △DEF is an equilateral triangle.

Hence, proved that △DEF is an equilateral triangle.

In the adjoining figure, ABCD is a quadrilateral in which AD = BC and P, Q, R, S are the mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.

Answer

By mid-point theorem,

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

In △ABD,

Since, P and Q are the mid-points of AB and BD respectively.

PQ || AD

⇒ PQ = $\dfrac{1}{2}$ × AD .....(1)

In △BCD,

Since, R and Q are the mid-points of DC and BD respectively.

QR || BC

⇒ QR = $\dfrac{1}{2}$ × BC

⇒ QR = $\dfrac{1}{2}$ × AD (∵ AD = BC) .....(2)

In △ABC,

Since, P and S are the mid-points of AB and AC respectively.

PS || BC

⇒ PS = $\dfrac{1}{2}$ × BC

⇒ PS = $\dfrac{1}{2}$ × AD (∵ AD = BC) .....(3)

In △ADC,

Since, S and R are the mid-points of AC and DC respectively.

SR || AD

⇒ SR = $\dfrac{1}{2}$ × AD .....(4)

From eq.(1), (2), (3) and (4), we have:

⇒ PQ = SR = QR = PS

Since, PQ || SR (Both are parallel to AD) and QR || PS (both are parallel to BC)

∴ PQRS is rhombus.

Hence, proved that PQRS is a rhombus.

In the adjoining figure, ABCD is a parallelogram in which E is the mid-point of DC and F is a point on AC such that CF = $\dfrac{1}{4}$ AC. If EF is produced to meet BC in G, prove that G is the mid-point of BC.

Answer

Join BD. Let BD intersect AC at point O.

We know that,

The diagonals of a parallelogram bisect each other.

AO = CO and OD = OB

Given,

⇒ CF = $\dfrac{1}{4}$ AC

⇒ CF = $\dfrac{1}{4}$ (AO + CO)

⇒ CF = $\dfrac{1}{4}$ (CO + CO)

⇒ CF = $\dfrac{1}{4}$ 2 CO

⇒ CF = $\dfrac{1}{2}$ CO

∴ F is the mid-point of CO.

By mid-point theorem,

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

In △COD,

Since, E and F are the mid-points of DC and OC respectively.

EF || OD

Since, EG and BD are straight lines. Thus, FG || OB

By 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.

In △COB,

Since, F is the mid-point of CO and FG || OB

∴ G is the mid-point of BC.

Hence, proved that G is the mid-point of BC.

In the adjoining figure, ABCD is a kite in which AB = AD and CB = CD. If E, F, G are respectively the mid-points of AB, AD and CD, prove that :

(i) ∠EFG = 90°

(ii) If GH || FE, then H bisects CB.

Answer

Join AC and BD, AC and BD intersects at O.

(i) We know that,

Diagonals of a kite intersect at right angles.

∠MON = 90° ...(1)

By mid-point theorem,

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.

In △ABD,

E and F are mid-points of AB and AD,

EF || BD and EF = $\dfrac{1}{2}$ BD

By converse of mid-point theorem,

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

In △ABO,

E is the mid-point of AB and EF || BD so, EM || BO,

∴ M is the mid-point of AO

⇒ AM = MO

In △AOD,

M and F are mid-points of AO and AD,

MF || OD and MF = $\dfrac{1}{2}$ OD ...(1)

In △ADC,

G and F are mid-points of CD and AD,

FG || AC and FG = $\dfrac{1}{2}$ AC (By mid-point theorem)

In △AOD,

F is the mid-point of AD and FG || AC so, FN || AO,

∴ N is the mid-point of OD (By converse of mid-point theorem)

⇒ ON = ND

From eq.(1), we have:

MF || OD and MF = $\dfrac{1}{2}$ OD

⇒ MF = ON

∴ OMFN is a parallelogram.

We know that,

Opposite angles of a parallelogram are equal.

⇒ ∠MON = ∠MFN = 90°

From figure,

∠EFG = ∠MFN = 90°

Hence, proved that ∠EFG = 90°.

(ii) EF || BD (Proved above)

A line through G is parallel to EF.

∴ GH || FE

or GH || BD

In △BCD,

GH || BD and G is mid-point of CD.

By converse of mid-point theorem,

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

∴ H is the mid-point of BC (By converse of mid-point theorem)

Hence, proved that if GH || FE, then H bisects CB.

In the adjoining figure, ABCD is a parallelogram, E is the mid-point of CD and through D, a line is drawn parallel to EB to meet CB produced at G and intersecting AB at F. Prove that :

(i) AD = $\dfrac{1}{2}$ GC

(ii) DG = 2 EB

Answer

By mid-point theorem,

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.

By converse of mid-point theorem,

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

Given, DF || EB

⇒ DG || EB

In △ DGC,

⇒ E is the mid-point of CD and DG || EB.

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

∴ BG = BC

Thus, BC = $\dfrac{1}{2}GC$

(i) We know that,

Opposite sides of a parallelogram are equal.

⇒ AD = BC

⇒ AD = $\dfrac{1}{2}$ GC

Hence, proved that AD = $\dfrac{1}{2}$ GC.

(ii) In △DGC,

Since, E and B are the mid-points of DC and GC respectively.

EB || DG

Thus, by mid-point theorem,

⇒ EB = $\dfrac{1}{2}$ DG

⇒ DG = 2 EB.

Hence, proved that DG = 2EB.

In the adjoining figure, in △ABC, AD is the median through A and E is the mid-point of AD. If BE produced meets AC in F, prove that AF = $\dfrac{1}{3}$ AC.

Answer

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

Given,

AD is the median to BC.

BD = CD

In △BCF,

Since, D is the mid-point of BC and DG // BF, thus by converse of mid-point theorem,

DG will bisect CF, thus G is the mid-point of CF.

⇒ CG = GF ......(1)

In △ADG,

Since, E is the mid-point of AD and EF // DG, thus by converse of mid-point theorem,

EF will bisect AG, thus F is the mid-point of AG.

⇒ AF = GF ........(2)

From (1) and (2),

⇒ AF = GF = CG

From figure,

⇒ AC = AF + GF + CG

⇒ AC = AF + AF + AF

⇒ AC = 3 AF

⇒ AF = $\dfrac{1}{3}$ AC.

Hence, proved that AF = $\dfrac{1}{3}$ AC.

In the adjoining figure, △ABC is right-angled at B and P is the mid-point of AC. Show that, PA = PB = PC.

Answer

Given,

P is mid-point of AC,

∴ PA = PC .........(1)

From figure,

PQ // CB

∠AQP = ∠ABC = 90° (Corresponding angles are equal)

⇒ ∠AQP + ∠PQB = 180° (Linear pair)

⇒ 90° + ∠PQB = 180°

⇒ ∠PQB = 180° - 90°

⇒ ∠PQB = 90°

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

In △ABC,

Since, P is the mid-point of AC and PQ // CB, thus by converse of mid-point theorem,

PQ will bisect AB, thus Q is the mid-point of AB.

In △AQP and △BQP,

⇒ ∠AQP = ∠PQB (Each equal to 90°)

⇒ PQ = PQ (Common side)

⇒ AQ = BQ (Q is the mid-point of AB)

Thus, △AQP ≅ △BQP. (By S.A.S. axiom)

∴ PB = PA (Corresponding parts of congruent triangles are equal) .........(2)

From (1) and (2), we get :

⇒ PB = PA = PC

Hence, proved that PA = PB = PC.

Show that the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a rhombus.

Answer

Join BD.

By mid-point theorem,

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.

In △ABC,

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

By mid-point theorem,

⇒ PQ || AC and PQ = $\dfrac{1}{2}$ AC ...(1)

In △ADC,

Since, S and R are the mid-points of AD and CD respectively.

⇒ SR || AC and SR = $\dfrac{1}{2}$ AC ...(2)

From eq.(1) and (2), we have :

PQ = SR and PQ || SR ...(3)

In △ABD,

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

PS || BD

⇒ PS = $\dfrac{1}{2}$ BD ...(4)

In △BCD,

Since, Q and R are the mid-points of BC and CD respectively.

QR || BD

⇒ QR = $\dfrac{1}{2}$ BD ...(5)

From eq.(4) and (5), we have:

PS = QR and PS || QR ...(6)

From eq.(3) and (6), we have:

In quadrilateral PQRS opposite sides are parallel and equal.

∴ PQRS is a parallelogram.

Given,

ABCD is a rectangle.

AB = CD and AD = BC

In △ASP and △BQP,

⇒ AP = BP (P is the mid-point of AB)

⇒ AS = BQ (As, S and Q are mid-points of equal sides AD and BC respectively)

⇒ ∠SAP = ∠QBP (Both equal to 90°)

∴ △ASP ≅ △BQP (By S.A.S axiom)

⇒ PS = PQ ....(7) (Corresponding parts of congruent traingles are equal)

From eq.(3), (6) and (7), we have:

PS = PQ = QR = SR

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

∴ PQRS is a rhombus.

Hence, the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a rhombus.

Show that the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rhombus is a rectangle.

Answer

By mid-point theorem,

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.

In △ABC,

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

PQ || AC

⇒ PQ = $\dfrac{1}{2}$ AC ...(1)

In △ADC,

Since, S and R are the mid-points of AD and CD respectively.

SR || AC

⇒ SR = $\dfrac{1}{2}$ AC ...(2)

From eq.(1) and (2), we have:

PQ = SR ...(3)

In △ABD,

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

PS || BD

⇒ PS = $\dfrac{1}{2}$ BD ...(4)

In △BCD,

Since, Q and R are the mid-points of BC and CD respectively.

QR || BD

⇒ QR = $\dfrac{1}{2}$ BD ...(5)

From eq.(4) and (5), we have:

PS = QR ...(6)

From eq.(3) and (6), we have:

∴ PQRS is a parallelogram.

We know that,

Diagonals of rhombus intersect at right angles.

⇒ ∠EOF = 90°

In quadrilateral OERF,

ER || OF and EO || RF

∴ OERF is a parallelogram.

Opposite angles of a parallelogram are equal.

⇒ ∠EOF = ∠ERF = 90°

In parallelogram PQRS,

⇒ ∠QRS = ∠QPS = 90°

⇒ ∠PSR = ∠PQR = x (let)

∠QRS + ∠QPS + ∠PSR + ∠PQR = 360°

⇒ 90° + 90° + x + x = 360°

⇒ 180° + 2x = 360°

⇒ 2x = 360° - 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$

⇒ x = 90°

⇒ ∠PSR = ∠PQR = 90°

Since, in parallelogram PQRS, opposite sides are equal and parallel and all the interior angles equal to 90°.

∴ PQRS is a rectangle.

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

Show that the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a square.

Answer

By mid-point theorem,

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.

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

We know that diagonals of a square are equal and bisect each other.

AC = BD

AO = OC = BO = OD

Join EF, FG, GH and HE.

Join AC and BD.

In △ACD,

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

By mid-point theorem,

∴ GH || AC and GH = $\dfrac{1}{2}$ AC

GH = AO (∵ O is the mid-point of AC) ...(1)

In △ABC,

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

∴ EF || AC and EF = $\dfrac{1}{2}$ AC

EF = AO (∵ O is the mid-point of AC) ...(2)

In △ABD,

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

∴ EH || BD and EH = $\dfrac{1}{2}$ BD

EH = BO (∵ O is the mid-point of BD)

∴ EH = AO ...(3)

In △BCD,

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

∴ FG || BD and FG = $\dfrac{1}{2}$ BD

FG = BO (∵ O is the mid-point of BD)

∴ FG = AO ....(4)

From eq.(1), (2), (3) and (4), we get:

EH || FG, EF || GH and EH = FG = GH = EF

Since, both the opposite sides of a quadrilateral are parallel.

∴ EFGH is a parallelogram.

In △GOH and △GOF,

⇒ OH = OF (Diagonals of parallelogram bisect each other)

⇒ OG = OG (Common side)

⇒ GH = GF (Proved above)

∴ △GOH ≅ △GOF (S.S.S axiom)

⇒ ∠GOH = ∠GOF (Corresponding parts of congruent triangles are equal)

From figure,

⇒ ∠GOH + ∠GOF = 180°

⇒ ∠GOH + ∠GOH = 180°

⇒ 2∠GOH = 180°

⇒ ∠GOH = $\dfrac{180°}{2}$

⇒ ∠GOH = 90°

So, the diagonals of EFGH bisect and are perpendicular to each other and all sides of quadrilateral EFGH are equal.

∴ EFGH is a square.

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

In the adjoining figure, ABCD is a trapezium in which AB || DC. If M and N are the mid-points of AC and BD respectively. Prove that MN = $\dfrac{1}{2}$ (AB - CD).

Answer

From figure,

AB // DC

EB // DC

In △BNE and △CND,

⇒ ∠BNE = ∠CND (Vertically opposite angles are equal)

⇒ ∠BEN = ∠NCD (Alternate angles are equal)

⇒ BN = DN (N is the mid-point of BD)

∴ △BNE ≅ △CND

⇒ BE = CD (Corresponding parts of congruent triangles are equal)

⇒ NE = CN (Corresponding parts of congruent triangles are equal)

∴ N is the mid-point of CE.

By mid-point theorem,

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.

Since, M and N are the mid-points of AC and CE respectively.

MN || AE

⇒ MN = $\dfrac{1}{2}$ AE

⇒ MN = $\dfrac{1}{2}$ (AB - BE)

⇒ MN = $\dfrac{1}{2}$ (AB - CD) (∵ BE = CD)

Hence, proved that MN = $\dfrac{1}{2}$ (AB - CD).

In the adjoining figure, ABCD is a trapezium in which AB || DC and E is the mid-point of AD. If EF || AB meets BC at F, show that F is the mid-point of BC.

Answer

Join AC. Let AC intersects EF at O.

Given,

AB || DC and EF || AB

∴ EF || AB || DC

By mid-point theorem,

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.

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

Since,

⇒ EF || DC

⇒ EO || DC

In △ADC,

E is the mid-point of AD and EO || DC.

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

Given,

⇒ EF || AB

⇒ OF || AB

In △ABC,

O is the mid-point of AC and OF || AB.

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

Hence, proved that F is the mid-point of BC.

Two points A and B lie on the same side of a line XY. If AD ⊥ XY and BE ⊥ XY meet XY in D and E respectively and C is the mid-point of AB, show that CD = CE.

Answer

Join BD which intersects CF at O.

Given,

CF ⊥ XY, AD ⊥ XY and BE ⊥ XY

⇒ CF || AD || BE

By mid-point theorem,

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.

By converse of mid-point theorem,

A line drawn through the midpoint of one side of a triangle, and parallel to another side, will bisect the third side.

Since,

⇒ CF || AD

⇒ CO || AD

In △ADB,

C is the mid-point of AB and CO || AD.

∴ O is the mid-point of BD. (By converse of mid-point theorem)

Given,

⇒ CF || BE

⇒ OF || BE

In △BDE,

O is the mid-point of BD and OF || BE.

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

⇒ DF = FE

In △CDF and △CEF,

⇒ DF = FE (Proved above)

⇒ CF = CF (Common side)

⇒ ∠CFD = ∠CFE (Both equal to 90°)

∴ △CDF ≅ △CEF (By S.A.S. axiom)

⇒ CD = CE (Corresponding parts of congruent triangles are equal)

Hence, proved that CD = CE.

Prove that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer

Let ABCD be the quadrilateral and E, F, G and H be the mid-point of AD, AB, BC and CD.

By mid-point theorem,

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.

Let ABCD be the quadrilateral and E, F, G and H are the mid-points of sides AD, AB, BC and CD.

In △BCD,

Since, G and H are the mid-points of BC and CD respectively.

⇒ GH || BD and GH = $\dfrac{1}{2}$ BD ...(1)

In △BAD,

Since, F and E are the mid-points of AB and AD respectively.

⇒ FE || BD and FE = $\dfrac{1}{2}$ BD ...(2)

From eq.(1) and (2), we have :

⇒ GH = FE and FE || GH

∴ EFGH is a parallelogram.

We know that,

Diagonals of the parallelogram, bisect each other.

EG and FH are the diagonals of parallelogram EFGH bisects each other.

Hence, proved that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.