Mid-point Theorem & its Converse [Including Intercept Theorem] — Test Yourself

The midpoints of the sides of a triangle are joined together to get four triangles. These four triangles are:

1. not equal to each other

2. congruent to each other

3. not congruent to each other

4. none of these

Answer

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 (By, mid-point theorem) or,

DE || FC ........(1)

DE || AF ........(2)

D and F are midpoints of AB and AC.

∴ DF || BC (By, mid-point theorem) or,

DF || EC ........(3)

DF || BE ......(4)

F and E are midpoints of AC and BC.

∴ FE || AB (By, mid-point theorem) or,

FE || AD .........(5)

FE || DB .........(6)

From (1) and (3) we get,

DE || FC and DF || EC.

Since, opposite sides of a parallelogram are parallel.

∴ DECF is a parallelogram

We know that,

Diagonal of || gm divides it into two congruent triangles.

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

∴ △DEF ≅ △ECF .........(7)

From (2) and (5) we get,

DE || AF and FE || AD.

Since, opposite sides of a parallelogram are parallel.

∴ ADEF is a parallelogram.

We know that,

Diagonal of || gm divides it into two congruent triangles.

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

∴ △DEF ≅ △AFD .........(8)

From (4) and (6) 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 .........(9)

From equations 7, 8 and 9 we get,

△AFD ≅ △DBE ≅ △ECF ≅ △DEF.

Thus, the four triangles formed are congruent to each other.

Hence, option 2 is the correct option.

In the given figure, AB || CD || EF. If AC = 7 cm, AE = 14 cm and BF = 20 cm, then DF is equal to:

1. 7 cm

2. 14 cm

3. 10 cm

4. 16 cm

Answer

Given, AB || CD || EF, AC = 7 cm, AE = 14 cm and BF = 20 cm.

Let the length of DF be x cm.

According to equal intercept theorem,

If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

From figure,

⇒ AE = AC + CE

⇒ 14 = 7 + CE

⇒ CE = 7 cm.

Since, OA makes equal intercept, so OB will also make equal intercepts.

∴ BD = DF = x (let)

From figure,

⇒ BF = BD + DF

⇒ 20 = x + x

⇒ 2x = 20

⇒ x = $\dfrac{20}{2}$ = 10 cm.

Hence, option 3 is the correct option.

In the given figure, AB || CD || EF and E is the mid-point of side AD, then :

1. OE : OF = 1 : 3

2. OE = OF

3. OF = 2 x OE

4. CF = FB

Answer

Given,

E is mid-point of AD.

Since, EF || DC ⇒ EO || DC

According to converse of the mid-point theorem, in any triangle, a straight line through the midpoint of one side that is parallel to a second side, then that line bisects the third side.

⇒ OA = OC

⇒ O is the mid-point of AC.

Since, EF || AB ⇒ OF || AB

According to converse of the mid-point theorem,

F will be the mif-point of BC.

⇒ CF = FB

Hence, option 4 is the correct option.

In rhombus PQRS; A, B and C are mid-points of sides PQ, QR and RS respectively. If ∠P = 60°, the angle PQR is equal to:

1. 60°

2. 90°

3. 120°

4. none of these

Answer

Given, ∠P = 60°

∠P and ∠PQR are consecutive angles in the rhombus.

The sum of consecutive angles in a rhombus is 180°.

⇒ ∠P + ∠PQR = 180°

⇒ 60° + ∠PQR = 180°

⇒ ∠PQR = 180° - 60°

⇒ ∠PQR = 120°.

Hence, option 3 is the correct option.

Statement 1: The diagonals of a quadrilateral are perpendicular to each other; P, Q, R and S are the midpoints of sides AB, BC, CD and DA respectively. Then PQRS will be a rectangle.

Statement 2: Quadrilateral PQRS will be a square as each of its angles will be 90°.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

From figure,

PA = PB ⇒ P is mid-point of AB

SA = SD ⇒ S is mid-point of AD

BQ = CQ ⇒ Q is mid-point of BC

CR = RD ⇒ R is mid-point of CD

So, statement 1 is true.

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,

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

⇒ PQ = $\dfrac{1}{2}$ AC and PQ || AC. [By mid-point theorem] ................(1)

In △ ADC,

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

⇒ SR = $\dfrac{1}{2}$ AC and SR || AC. [By mid-point theorem] .................(2)

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

PQ = SR and PQ || SR.

In △ BCD,

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

⇒ QR = $\dfrac{1}{2}$ BD and QR || BD. [By mid-point theorem] .................(3)

In △ ABD,

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

⇒ PS = $\dfrac{1}{2}$ BD and PS || BD. [By mid-point theorem] .................(4)

From (3) and (4), we get :

QR = PS and QR || PS.

Since, diagonals of quadrilateral intersect at right angle.

∴ ∠AOD = ∠COD = ∠AOB = ∠BOC = 90°.

From figure,

PQ || AC

∴ ∠PXO = ∠AOD = 90° (Corresponding angles are equal)

∴ ∠QXO = ∠COD = 90° (Corresponding angles are equal)

SR || AC

∴ ∠SZO = ∠AOB = 90° (Corresponding angles are equal)

∴ ∠RZO = ∠BOC = 90° (Corresponding angles are equal)

PS || BD

∴ ∠S = ∠RZO = 90° (Corresponding angles are equal)

∴ ∠P = ∠QXO = 90° (Corresponding angles are equal)

QR || BD

∴ ∠R = ∠SZO = 90° (Corresponding angles are equal)

∴ ∠Q = ∠PXO = 90° (Corresponding angles are equal)

Since, in quadrilateral PQRS,

Each interior angle is equal to 90° and opposite sides are parallel and equal, and we can prove that PQRS is a rectangle but cannot prove that all sides are equal thus we cannot prove that it is a square.

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Statement 1: AD is median of triangle ABC and DE is parallel to BA. Then DE will bisect AC.

Statement 2: DE is median of triangle ADC.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

In △ ABC,

From figure,

BD = DC.

∴ D is the mid-point of BC.

A median of a triangle is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

So, AD is the median of triangle ABC.

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.

Since, D is mid-point of BC and AB || DE

∴ E is the mid-point of AC.

Thus, DE is a median of triangle ADC.

Thus, DE will bisect AC.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): The figure formed by joining the mid-points of the sides of a quadrilateral ABCD is a square.

Reason (R): Diagonals of quadrilateral ABCD are not equal and are not perpendicular to each other.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

For the figure formed by joining the midpoints of a quadrilateral to be a square, the diagonals of the original quadrilateral must be both equal and perpendicular to each other.

The reason states that the diagonals are not equal and not perpendicular.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Assertion (A): R, S, D and E are mid-points of OC, OB, AB and AC respectively, then DERS is a parallelogram.

Reason (R): DS ∥ AO ∥ ER and DS = ER = $\dfrac{1}{2}AO$.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

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 △ABO,

D and S are respective midpoints of AB and BO.

∴ DS || AO and DS = $\dfrac{1}{2}$ AO [By mid-point theorem].................(1)

In △ACO,

E and R are respective midpoints of AC and CO.

∴ ER || AO and ER = $\dfrac{1}{2}$ AO [By mid-point theorem]..................(2)

From (1) and (2) we get,

DS || ER and DS = ER = $\dfrac{1}{2}$ AO

So, reason (R) is true.

We know that,

If one pair of opposite sides of a quadrilateral are equal in length and parallel, then the quadrilateral is a parallelogram.

∴ DERS is a parallelogram.

So, assertion (A) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

In triangle ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm; find the perimeter of the parallelogram BDEF.

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 of a triangle parallel to another, bisects the third side.

Given,

E is mid-point of AC and EF || AB.

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

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

∴ DE || BC and DE = $\dfrac{1}{2}BC$ (By mid-point theorem)

⇒ DE || BF and DE = BF (As F is mid-point of BC).

Given,

EF || AB

∴ EF || BD.

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

∴ EF = $\dfrac{1}{2}AB$ = BD. (By mid-point theorem)

Since, opposite sides of quadrilateral BDEF are parallel and equal.

∴ BDEF is a parallelogram.

From figure,

⇒ BD = $\dfrac{1}{2}AB = \dfrac{1}{2} \times 8$ = 4 cm,

⇒ BF = $\dfrac{1}{2}BC = \dfrac{1}{2} \times 9$ = 4.5 cm.

Perimeter of BDEF = BD + DE + EF + BF

= BD + BF + BD + BF (Since opposite sides of parallelogram are equal)

= 4 + 4.5 + 4 + 4.5

= 17 cm.

Hence, perimeter of parallelogram BDEF = 17 cm.

P, Q and R are mid-points of sides AB, BC and CD respectively of a rhombus ABCD. Show that PQ is perpendicular to QR.

Answer

Join diagonals of rhombus AC and BD.

We know that,

Diagonals of rhombus intersect at 90°.

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,

P and Q are mid-points of sides AB and BC respectively.

∴ PQ || AC (By mid-point theorem)

In △ BCD,

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

∴ QR || BD (By mid-point theorem)

Since, AC ⊥ BD and PQ || AC and QR || BD.

∴ PQ ⊥ QR.

Hence, PQ is perpendicular to QR.

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral obtained by joining the mid-points of its adjacent sides 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.

Let ABCD be a quadrilateral where P, Q, R and S are the mid-point of AB, BC, CD and DA.

In △ ABC,

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

⇒ PQ = $\dfrac{1}{2}AC$ and PQ || AC. [By mid-point theorem] .......(1)

In △ ADC,

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

⇒ SR = $\dfrac{1}{2}AC$ and SR || AC. [By mid-point theorem] .......(2)

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

PQ = SR and PQ || SR.

In △ BCD,

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

⇒ QR = $\dfrac{1}{2}BD$ and QR || BD. [By mid-point theorem] .......(3)

In △ ABD,

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

⇒ PS = $\dfrac{1}{2}BD$ and PS || BD. [By mid-point theorem] .......(4)

From (3) and (4), we get :

QR = PS and QR || PS.

Since, diagonals of quadrilateral intersect at right angle.

∴ ∠AOD = ∠COD = AOB = ∠BOC = 90°.

From figure,

PQ || AC

∴ ∠PXO = ∠AOD = 90° (Corresponding angles are equal)

∴ ∠QXO = ∠COD = 90° (Corresponding angles are equal)

SR || AC

∴ ∠SZO = ∠AOB = 90° (Corresponding angles are equal)

∴ ∠RZO = ∠BOC = 90° (Corresponding angles are equal)

PS || BD

∴ ∠S = ∠RZO = 90° (Corresponding angles are equal)

∴ ∠P = ∠QXO = 90° (Corresponding angles are equal)

QR || BD

∴ ∠R = ∠SZO = 90° (Corresponding angles are equal)

∴ ∠Q = ∠PXO = 90° (Corresponding angles are equal)

Since, in quadrilateral PQRS,

Each interior angle is equal to 90° and opposite sides are parallel and equal.

∴ PQRS is a rectangle.

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

In △ ABC, E is mid-point of the median AD and BE produced meets side AC at point Q. Show that BE : EQ = 3 : 1.

Answer

Draw DY || BQ.

In △ BCQ and △ DCY,

⇒ ∠BCQ = ∠DCY (Common)

⇒ ∠BQC = ∠DYC (Corresponding angles are equal)

∴ △ BCQ ~ △ DCY (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{BQ}{DY} = \dfrac{BC}{DC} = \dfrac{CQ}{CY}$ ..........(1)

Since, D is the mid-point of BC.

∴ BC = 2CD

Considering L.H.S. of the equation (1), we get :

$\Rightarrow \dfrac{BQ}{DY} = \dfrac{2CD}{DC} \\[1em] \Rightarrow \dfrac{BQ}{DY} = 2 \text{ ........(1)}$

In △ AEQ and △ ADY,

⇒ ∠EAQ = ∠DAY (Common)

⇒ ∠AEQ = ∠ADY (Corresponding angles are equal)

∴ △ AEQ ~ △ ADY (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{EQ}{DY} = \dfrac{AE}{AD} = \dfrac{1}{2}$ (Since, E is the mid-point of AD)

$\Rightarrow \dfrac{EQ}{DY} = \dfrac{1}{2}$ ............(2)

Dividing equation (1) by (2), we get :

$\Rightarrow \dfrac{\dfrac{BQ}{DY}}{\dfrac{EQ}{DY}} = \dfrac{2}{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{BQ \times DY}{EQ \times DY} = 4 \\[1em] \Rightarrow \dfrac{BQ}{EQ} = 4 \\[1em] \Rightarrow \dfrac{BE + EQ}{EQ} = 4 \\[1em] \Rightarrow BE + EQ = 4EQ \\[1em] \Rightarrow BE = 4EQ - EQ \\[1em] \Rightarrow BE = 3EQ \\[1em] \Rightarrow \dfrac{BE}{EQ} = \dfrac{3}{1}.$

Hence, proved that BE : EQ = 3 : 1.

In the given figure, M is the mid-point of AB and DE, whereas N is mid-point of BC and DF. Show that : EF = AC.

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 △ EDF,

M is the mid-point of ED and N is the mid-point of DF.

∴ MN = $\dfrac{1}{2}EF$ (By mid-point theorem)

⇒ EF = 2MN .............(1)

In △ ABC,

M is the mid-point of AB and N is the mid-point of BC.

∴ MN = $\dfrac{1}{2}AC$ (By mid-point theorem)

⇒ AC = 2MN .............(2)

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

⇒ EF = AC.

Hence, proved that EF = AC.

In triangle ABC; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm, find the perimeter of the parallelogram BDEF.

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 of a triangle parallel to another, bisects the third side.

Given,

E is mid-point of AC and EF || AB.

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

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

∴ DE || BC and DE = $\dfrac{1}{2}BC$ (By mid-point theorem)

⇒ DE || BF and DE = BF (As F is mid-point of BC).

Given,

EF || BC

∴ EF || BD.

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

∴ EF = $\dfrac{1}{2}AB$ = BD. (By mid-point theorem)

Since, opposite sides of quadrilateral BDEF are parallel and equal.

∴ BDEF is a parallelogram.

From figure,

⇒ BD = $\dfrac{1}{2}AB = \dfrac{1}{2} \times 16$ = 8 cm,

⇒ BF = $\dfrac{1}{2}BC = \dfrac{1}{2} \times 18$ = 9 cm.

Perimeter of BDEF = BD + DE + EF + BF

= BD + BF + BD + BF (Since opposite sides of parallelogram are equal)

= 8 + 9 + 8 + 9

= 34 cm.

Hence, perimeter of parallelogram BDEF = 34 cm.

In the given figure, AD and CE are medians and DF // CE. Prove that : FB = $\dfrac{1}{4}AB$.

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 of a triangle parallel to another, bisects the third side.

Since, AD and CE are medians.

∴ D is the mid-point of BC and E is the mid-point of AB.

In △ BEC,

DF || CE and D is the mid-point of BC.

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

∴ FB = $\dfrac{1}{2}BE$ .......(1)

Since, E is the mid-point of AB.

∴ BE = $\dfrac{1}{2}AB$ .......(2)

Substituting value of BE from equation (2) in (1), we get :

∴ FB = $\dfrac{1}{2} \times \dfrac{1}{2} \times AB = \dfrac{1}{4}AB$.

Hence, proved that FB = $\dfrac{1}{4}AB$.

In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P. Prove that :

(i) BP = 2AD

(ii) O is mid-point of AP.

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 of a triangle parallel to another, bisects the third side.

(i) In △ ABP,

⇒ E is the mid-point of AB and EC || AP.

∴ C is the mid-point of BP. (By converse of mid-point theorem)

∴ BP = 2BC .........(1)

Since, ABCD is a parallelogram.

∴ AD = BC (Opposite sides of parallelogram are equal) .......(2)

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

⇒ BP = 2AD.

Hence, proved that BP = 2AD.

(ii) Since, opposite sides of parallelogram are parallel.

∴ AB || CD

⇒ AB || OC.

In △ ABP,

⇒ E is the mid-point of AB and OC || AB.

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

Hence, O is the mid-point of AP.

In a trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.

Prove that :

AB + DC = 2EF.

Answer

Join BE and produce to meet CD produced at point P.

In △ PDE and △ BAE,

⇒ ∠PED = ∠BEF (Vertically opposite angles are equal)

⇒ AE = ED (Since, E is the mid-point of AD)

⇒ ∠EDP = ∠EAB (Alternate angles are equal)

∴ △ PDE ≅ △ BAE (By A.S.A. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BE = EP and AB = PD.

In △ BPC,

Since, E and F are mid-points of sides BP and BC respectively.

∴ EF = $\dfrac{1}{2}PC$.

To prove :

AB + CD = 2EF ........(1)

Substituting value in L.H.S. of equation (1), we get :

⇒ AB + CD = PD + CD = PC.

Substituting value in R.H.S. of equation (2), we get :

⇒ 2EF = $2 \times \dfrac{1}{2}PC$ = PC.

Since, L.H.S. = R.H.S.

Hence, proved that AB + CD = 2EF.

In △ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.

Answer

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 △ ABC,

Since, AD is the median.

∴ D is the mid-point of BC.

Since, D is mid-point of BC and DE || AB.

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

Join BE.

Hence, proved that BE is also a median.

Adjacent sides of a parallelogram are equal and one of diagonals is equal to any one of the sides of this parallelogram. Show that its diagonals are in the ratio $\sqrt{3} : 1$.

Answer

Let ABCD be the required parallelogram.

∴ AB = CD and BC = AD. (Opposite sides of parallelogram are equal)

Given,

Adjacent sides of a parallelogram are equal.

∴ AB = BC.

∴ AB = BC = CD = AD

Since, all sides of parallelogram are equal.

∴ ABCD is a rhombus.

Given, one of the diagonals is equal to its sides. Let diagonal BD be equal to sides.

∴ AB = BC = CD = AD = BD = a (let).

From figure,

⇒ BO = $\dfrac{BD}{2} = \dfrac{a}{2}$ (Since, in a rhombus diagonals bisect each other at right angle).

Hence, △ AOB is right-angled at O.

In △ AOB,

By pythagoras theorem,

⇒ AB² = BO² + AO²

⇒ a² = $\Big(\dfrac{a}{2}\Big)^2$ + AO²

⇒ AO² = $a^2 - \dfrac{a^2}{4}$

⇒ AO² = $\dfrac{4a^2 - a^2}{4}$

⇒ AO² = $\dfrac{3a^2}{4}$

⇒ AO = $\dfrac{\sqrt{3}a}{2}$,

⇒ AC = 2AO = $2 \times \dfrac{\sqrt{3}a}{2} = \sqrt{3}a$ units.

The ratio of the diagonals is:

$\Rightarrow \dfrac{AC}{BD} = \dfrac{\sqrt{3}a}{a} = \dfrac{\sqrt{3}}{1}$

∴ AC : BD = $\sqrt{3}$ : 1.

Hence, proved that diagonals are in the ratio $\sqrt{3}$ : 1.