Mid-point Theorem & its Converse [Including Intercept Theorem] — Exercise 11(A)

In the given figure, ABCD is a rectangle. As per the given information, the length of PQ is :

1. 12 cm

2. 14 cm

3. 20 cm

4. 10 cm

Answer

From figure,

AP = PB and BQ = QC.

∴ P is the mid-point of AB and Q is the mid-point of BC.

In △ ADC,

⇒ AD² + CD² = AC² (By pythagoras theorem)

⇒ 12² + 16² = AC²

⇒ 144 + 256 = AC²

⇒ AC² = 400

⇒ AC = $\sqrt{400}$ = 20 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 is equal to half of it.

∴ PQ = $\dfrac{1}{2} AC = \dfrac{1}{2} \times 20$ = 10 cm.

Hence, Option 4 is the correct option.

The quadrilateral obtained by joining the mid-points (in order) of the sides of quadrilateral ABCD is :

1. rectangle

2. rhombus

3. parallelogram

4. square

Answer

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

Join PQRS, AC and 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,

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

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

In △ ADC,

S and R are mid-points of sides AD and DC respectively.

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

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

⇒ PQ = SR and PQ || SR.

In △ ABD,

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

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

In △ CBD,

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

∴ QR || BD and QR = $\dfrac{1}{2}BD$ (By mid-point theorem) ........(4)

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

⇒ SP = QR and SP || QR.

Since, opposite sides are parallel and equal.

∴ PQRS is a parallelogram.

Hence, Option 3 is the correct option.

If BC = 12 cm, AB = 14.8 cm, AC = 12.8 cm, the perimeter of quadrilateral BCYX is :

1. 31.8 cm

2. 15.9 cm

3. 29.8 cm

4. 32.8 cm

Answer

From figure,

X and Y are the mid-point of AB and AC respectively.

∴ BX = $\dfrac{AB}{2} = \dfrac{14.8}{2}$ = 7.4 cm and CY = $\dfrac{AC}{2} = \dfrac{12.8}{2}$ = 6.4 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 is equal to half of it.

In △ ABC,

∴ XY = $\dfrac{1}{2}BC = \dfrac{1}{2} \times 12$ = 6 cm.

Perimeter of BCYX = BC + CY + XY + BX = 12 + 6.4 + 6 + 7.4 = 31.8 cm

Hence, Option 1 is the correct option.

In the given figure, AB = AC, P, Q and R are mid-points of sides BC, CA and AB respectively, then △ PQR is :

1. scalene

2. isosceles

3. equilateral

4. obtuse angled

Answer

Given,

AB = AC = x (let)

From figure,

AR = RB, BP = PC and AQ = QC.

∴ R, P and Q are mid-points of sides AB, BC and AC respectively.

Join QR.

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.

∴ PQ = $\dfrac{1}{2}AB = \dfrac{x}{2}$, PR = $\dfrac{1}{2}AC = \dfrac{x}{2}$ and QR = $\dfrac{1}{2}BC$.

In △ PQR,

PQ = PR.

∴ △ PQR is an isosceles triangle.

Hence, Option 2 is the correct option.

P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively of rectangle ABCD, then quadrilateral PQRS is :

1. rectangle

2. rhombus

3. square

4. parallelogram

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.

Join AC and BD.

We know that,

Diagonals of a rectangle are equal.

∴ AC = BD = x (let)

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively.

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

In Δ ADC, S and R are mid-points of sides AD and CD respectively.

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

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

PQ || SR and PQ = SR

In Δ ABD, P and S are the mid-points of sides AB and AD respectively.

∴ PS || BD and PS = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By mid-point theorem) ........(3)

In Δ BDC, Q and R are mid-points of sides BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By mid-point theorem) .........(4)

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

PQ || SR and PS = QR

By using equation (1), (2), (3) and (4), we get :

⇒ PQ = QR = SR = PS

∴ PQRS is a rhombus.

Hence, Option 2 is the correct option.

In triangle ABC, M is the mid-point of AB and a straight line through M and parallel to BC cuts AC at N. Find the lengths of AN and MN, if BC = 7 cm and AC = 5 cm.

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.

∴ N bisects AC.

∴ AN = $\dfrac{AC}{2} = \dfrac{5}{2}$ = 2.5 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 is equal to half of it.

∴ MN = $\dfrac{1}{2}BC = \dfrac{1}{2} \times 7$ = 3.5 cm.

Hence, AN = 2.5 cm and MN = 3.5 cm.

Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.

Answer

Let ABCD be the rectangle and P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively. Join PQRS.

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.

Join AC and BD.

We know that,

Diagonals of a rectangle are equal.

∴ AC = BD = x (let)

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively.

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

In Δ ADC, S and R are mid-points of sides AD and CD respectively.

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

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

PQ || SR and PQ = SR

In Δ ABD, P and S are the mid-points of sides AB and AD respectively.

∴ PS || BD and PS = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By mid-point theorem) ........(3)

In Δ BDC, Q and R are mid-points of sides BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By mid-point theorem) .........(4)

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

PQ || SR and PS = QR

By using equation (1), (2), (3) and (4), we get :

⇒ PQ = QR = SR = PS

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

∴ PQRS is a rhombus.

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

D, E and F are the mid-points of the sides AB, BC and CA of an isosceles △ ABC in which AB = BC. Prove that △ DEF is also isosceles.

Answer

Join D, E and F.

Given,

AB = BC = x (let)

Given,

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

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.

∴ DF = $\dfrac{1}{2}BC = \dfrac{x}{2}$, FE = $\dfrac{1}{2}AB = \dfrac{x}{2}$ and DE = $\dfrac{1}{2}AC$.

In △ DEF,

DF = FE.

∴ △ DEF is an isosceles triangle.

Hence, proved that DEF is an isosceles triangle.

The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that :

PR = $\dfrac{1}{2}$ (AB + CD)

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,

⇒ PR // AB

⇒ PQ // AB

In △ ABD,

P is mid-point of AD and PQ // AB.

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

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

Given,

⇒ PR // DC

⇒ QR // DC

In △ BCD,

Q is mid-point of BD and QR // DC.

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

∴ QR = $\dfrac{1}{2}CD$ (By mid-point theorem) ..........(2)

Adding equations (1) and (2), we get :

⇒ PQ + QR = $\dfrac{1}{2}AB + \dfrac{1}{2}CD$

⇒ PR = $\dfrac{1}{2}(AB + CD)$.

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

The figure, given below, shows a trapezium ABCD. M and N are the mid-points of the non-parallel sides AD and BC respectively. Find :

(i) MN, if AB = 11 cm and DC = 8 cm.

(ii) AB, if DC = 20 cm and MN = 27 cm.

(iii) DC, if MN = 15 cm and AB = 23 cm.

Answer

Join BD. Let BD intersects MN at point O.

We know that,

In trapezium the line joining the mid-points of non-parallel sides are parallel to the parallel sides of trapezium.

∴ MN || 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,

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

Given,

⇒ MN || AB

⇒ MO || AB

In △ ABD,

M is mid-point of AD and MO || AB.

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

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

Given,

⇒ MN || DC

⇒ ON || DC

In △ BCD,

O is mid-point of BD and N is mid-point of BC.

∴ ON = $\dfrac{1}{2}CD$ (By mid-point theorem) ..........(2)

Adding equations (1) and (2), we get :

⇒ MO + ON = $\dfrac{1}{2}AB + \dfrac{1}{2}CD$

⇒ MN = $\dfrac{1}{2}(AB + CD)$ ..........(3)

(i) Given,

⇒ AB = 11 cm

⇒ DC = 8 cm

Substituting values in equation (3), we get :

$\Rightarrow MN = \dfrac{1}{2}(AB + CD) \\[1em] \Rightarrow MN = \dfrac{1}{2} \times (11 + 8) \\[1em] \Rightarrow MN = \dfrac{1}{2} \times 19 \\[1em] \Rightarrow MN = 9.5 \text{ cm}.$

Hence, MN = 9.5 cm

(ii) Given,

⇒ MN = 27 cm

⇒ DC = 20 cm

Substituting values in equation (3), we get :

$\Rightarrow MN = \dfrac{1}{2}(AB + CD) \\[1em] \Rightarrow 27 = \dfrac{1}{2} \times (AB + 20) \\[1em] \Rightarrow 27 \times 2 = AB + 20 \\[1em] \Rightarrow AB + 20 = 54 \\[1em] \Rightarrow AB = 54 - 20 = 34 \text{ cm}.$

Hence, AB = 34 cm.

(iii) Given,

⇒ MN = 15 cm

⇒ AB = 23 cm

Substituting values in equation (3), we get :

$\Rightarrow MN = \dfrac{1}{2}(AB + CD) \\[1em] \Rightarrow 15 = \dfrac{1}{2} \times (23 + DC) \\[1em] \Rightarrow 15 \times 2 = 23 + DC \\[1em] \Rightarrow 23 + DC = 30 \\[1em] \Rightarrow DC = 30 - 23 = 7 \text{ cm}.$

Hence, DC = 7 cm.

The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral 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 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.

L and M are the mid-points of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.

Answer

From figure,

As, ABCD is a parallelogram.

∴ AB = CD

⇒ $\dfrac{AB}{2} = \dfrac{CD}{2}$

⇒ BL = DM

As, AB || CD

∴ BL || DM

Since, in quadrilateral DLBM,

BL = DM and BL || DM

∴ DLBM is a parallelogram.

So we get, DL parallel to MB.

Suppose DL and BM intersect AC at P and Q respectively.

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 triangle ABQ,

⇒ PL || QB (As, DL || MB)

⇒ Given, L is the midpoint of AB.

∴ P is the mid point of AQ (By converse of mid-point theorem)

∴ AP = PQ ........(1)

In triangle CDP,

⇒ QM || PD (As, BM || DL)

⇒ Given, M is the midpoint of CD.

∴ Q is the mid point of CP (By converse of mid-point theorem)

∴ PQ = CQ ........(2)

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

⇒ AP = PQ = CQ.

Hence, proved that DL and BM trisect the diagonal AC.

ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and AC respectively. Prove that EFGH 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 is equal to half of it.

In △ ADC,

G and H are mid-points of sides CD and AC respectively.

⇒ GH = $\dfrac{1}{2}AD$

⇒ AD = 2GH .........(1)

In △ ABD,

E and F are mid-points of sides AB and BD respectively.

⇒ EF = $\dfrac{1}{2}AD$

⇒ AD = 2EF .........(2)

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

⇒ AD = 2GH = 2EF ........(3)

In △ BCD,

G and F are mid-points of sides CD and BD respectively.

⇒ GF = $\dfrac{1}{2}BC$

⇒ BC = 2GF .........(4)

In △ ABC,

E and H are mid-points of sides AB and AC respectively.

⇒ EH = $\dfrac{1}{2}BC$

⇒ BC = 2EH .........(5)

From equations (4) and (5), we get :

⇒ BC = 2GF = 2EH ........(6)

Given,

⇒ AD = BC .............(7)

From equations (3), (6) and (7), we get :

⇒ 2GH = 2EF = 2GF = 2EH

⇒ GH = EF = GF = EH.

Since, all sides of quadrilateral EFGH are equal.

∴ EFGH is a rhombus.

Hence, proved that EFGH is a rhombus.

A parallelogram ABCD has P the mid-point of DC and Q a point of AC such that CQ = $\dfrac{1}{4}AC$. PQ produced meets BC at R.

Prove that :

(i) R is the mid-point of BC,

(ii) PR = $\dfrac{1}{2}DB$.

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.

We know that,

Diagonals of parallelogram bisect each other.

∴ AX = CX and BX = DX.

Given,

⇒ CQ = $\dfrac{1}{4}AC$

⇒ CQ = $\dfrac{1}{4} \times (AX + CX)$

⇒ CQ = $\dfrac{1}{4} \times (CX + CX) = \dfrac{1}{4} \times 2CX = \dfrac{CX}{2}$.

∴ Q is the mid-point of CX.

(i) In △ CDX,

P and Q are mid-point of sides CD and CX.

∴ PQ || DX (By mid-point theorem)

Since, DXB and PQR are straight line.

∴ PR || DB

∴ QR || XB.

In △ CXB,

Q is the mid-point of CX and QR || XB.

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

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

(ii) In △ BCD,

P and R are mid-point of sides CD and BC.

∴ PR || BD and PR = $\dfrac{1}{2}BD$.

Hence, proved that PR = $\dfrac{1}{2}BD$.