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

In the given figure, l // m // n and D is mid-point of CE. If AE = 12.6 cm, then BD is :

1. 12.6 cm

2. 25.2 cm

3. 6.3 cm

4. 18.9 cm

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.

In △ AEC,

Given,

⇒ m // n

∴ BD // AE.

D is mid-point of CE.

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

Also,

⇒ BD = $\dfrac{1}{2}AE = \dfrac{1}{2} \times 12.6$ = 6.3 cm

Hence, Option 3 is the correct option.

In a trapezium ABCD, AB // DC, E is mid-point of AD and F is mid-point of BC, then :

1. 2EF = $\dfrac{1}{2}(AB + DC)$

2. 2EF = AB + DC

3. EF = AB + DC

4. EF = $\dfrac{1}{2} \times AB \times DC$

Answer

Join AC. Let AC intersects EF 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.

∴ AB || EF || 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,

⇒ EF || DC

⇒ EO || DC

In △ ADC,

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

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

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

Given,

⇒ EF || AB

⇒ OF || AB

In △ ABC,

O is mid-point of AC and F is mid-point of BC.

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

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

⇒ EO + OF = $\dfrac{1}{2}DC + \dfrac{1}{2}AB$

⇒ EF = $\dfrac{1}{2}(DC + AB)$

⇒ 2EF = AB + DC.

Hence, Option 2 is the correct option.

The given figure shows a parallelogram ABCD in which E is mid-point of AD and DL // EB. Then, BF is equal to :

1. AD

2. BE

3. AE

4. AB

Answer

By 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.

In parallelogram ABCD,

DL || EB

Since, E is mid-point of AD.

∴ AE = ED

∴ BL = LC (By equal intercept theorem)

In △ BLF and △ DLC,

⇒ ∠BLF = ∠DLC (Vertically opposite angles are equal)

⇒ BL = LC (Proved above)

⇒ ∠LBF = ∠LCD (Alternate angles are equal)

∴ △ BLF ≅ △ DLC (By A.S.A. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BF = CD .........(1)

We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD .........(2)

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

⇒ BF = AB.

Hence, Option 4 is the correct option.

In the given figure AD and BE are medians, then ED is equal to :

1. 2AB

2. $\dfrac{1}{2}AB$

3. $\dfrac{1}{4}AB$

4. $\dfrac{1}{8}AB$

Answer

Join ED.

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, AD and BE are medians.

∴ D is mid-point of BC and E is mid-point of AC.

In △ ABC,

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

Hence, Option 2 is the correct option.

In the quadrilateral ABCD, if AB // CD, E is mid-point of side AD and F is mid-point of BC. If AB = 20 cm and EF = 16 cm, the length of side DC is :

1. 18 cm

2. 12 cm

3. 24 cm

4. 32 cm

Answer

Join BD. Let BD intersect EF 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.

∴ AB || EF || 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,

⇒ EF || AB

⇒ EO || AB

In △ ABD,

E is mid-point of AD and EO || AB.

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

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

Given,

⇒ EF || DC

⇒ OF || DC

In △ BCD,

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

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

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

⇒ EO + OF = $\dfrac{1}{2}AB + \dfrac{1}{2}CD$

⇒ EF = $\dfrac{1}{2}(AB + CD)$

Substituting values we get :

$\Rightarrow 16 = \dfrac{1}{2}(20 + CD) \\[1em] \Rightarrow 16 \times 2 = 20 + CD \\[1em] \Rightarrow 32 = 20 + CD \\[1em] \Rightarrow CD = 32 - 20 = 12\text{ cm}.$

Hence, Option 2 is the correct option.

Use the following figure to find :

(i) BC, if AB = 7.2 cm.

(ii) GE, if FE = 4 cm.

(iii) AE, if BD = 4.1 cm.

(iv) DF, if CG = 11 cm.

Answer

By 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.

(i) From figure,

CG || BF || AE

Since, CD = DE

∴ BC = AB = 7.2 cm (By equal intercept theorem)

Hence, BC = 7.2 cm.

(ii) From figure,

CG || BF || AE

Since, CD = DE

∴ FG = FE = 4 cm (By equal intercept theorem)

From figure,

⇒ GE = FG + FE = 4 + 4 = 8 cm.

Hence, GE = 8 cm.

(iii) 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.

In △ AEC,

D is mid-point of CE

⇒ BF || AE

∴ BD || AE.

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

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

⇒ AE = 2BD = 2 × 4.1 = 8.2 cm

Hence, AE = 8.2 cm.

(iv) In △ EGC,

D is mid-point of CE.

⇒ BF || CG

∴ DF || CG.

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

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

⇒ DF = $\dfrac{1}{2} \times 11$ = 5.5 cm

Hence, DF = 5.5 cm.

In the figure, given below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that :

(i) AQ // BS

(ii) DS = 3RS

Answer

Given,

P is mid-point of AB.

∴ AP = PB

Since, 2AD = AB

∴ AD = $\dfrac{AB}{2}$

∴ AP = AB = AD.

(i) 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 △ DPR,

A and Q are mid-points of DP and DR respectively.

∴ AQ || PR [By mid-point theorem] .......(1)

Given,

⇒ PR || BS ...........(2)

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

⇒ AQ || BS.

Hence, proved that AQ || BS.

(ii) By 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,

PR || BS

Since, AD = AP = PB

∴ DQ = QR = RS .........(3)

From figure,

⇒ DS = DQ + QR + RS

⇒ DS = RS + RS + RS [From equation (3)]

⇒ DS = 3RS.

Hence, proved that DS = 3RS.

The side AC of a triangle ABC is produced to point E so that CE = $\dfrac{1}{2}AC$. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that :

(i) 3DF = EF

(ii) 4CR = 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.

By 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.

(i) In △ ABC,

D is the mid-point of BC and DP || AB.

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

∴ AP = PC

Given,

⇒ CE = $\dfrac{1}{2}AC$

∴ CE = PC

Since, AP = PC and CE = PC,

∴ AP = PC = CE.

Since,

⇒ AB || DP || CR

⇒ AF || DP || CR (Since, point F lies on straight line AB)

In △ AEF,

AF || PD || CR and AP = PC = CE

∴ DF = DR = RE = x (let) [By equal intercept theorem]

From figure,

⇒ EF = DF + DR + RE = x + x + x = 3x = 3DF.

Hence, proved that 3DF = EF.

(ii) In △ ABC,

D is mid-point of BC and DP || AB.

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

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

In △ PED,

Given,

⇒ CE = $\dfrac{1}{2}AC$

⇒ CE = PC (Since, P is mid-point of AC)

∴ C is the mid-point of PE.

C is mid-point of PE and DP || CR.

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

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

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

⇒ CR = $\dfrac{1}{2} \times \dfrac{1}{2}AB$

⇒ CR = $\dfrac{1}{4}AB$

⇒ 4CR = AB.

Hence, proved that 4CR = AB.

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :

(i) M, A and N are collinear.

(ii) A is the mid-point of MN.

Answer

In △ AQN and △ BQC,

⇒ AQ = BQ (Since, CQ is the median)

⇒ QN = CQ (Given)

⇒ ∠AQN = ∠CQB (Vertically opposite angles are equal)

∴ △ AQN ≅ △ BQC (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ ∠QAN = ∠QBC ........(1)

⇒ BC = AN ..........(2)

In △ APM and △ CPB,

⇒ AP = CP (Since, BP is the median)

⇒ PM = BP (Given)

⇒ ∠APM = ∠CPB (Vertically opposite angles are equal)

∴ △ APM ≅ △ CPB (By S.A.S. axiom)

⇒ ∠PAM = ∠PCB [By C.P.C.T.C.] ........(3)

⇒ BC = AM [By C.P.C.T.C.] ..........(4)

(i) In △ ABC,

By angle sum property of triangle,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ ∠QBC + ∠PCB + ∠BAC = 180°

⇒ ∠QAN + ∠PAM + ∠BAC = 180° [From equations (1) and (3)]

Since, the sum of above angles equal to 180°.

∴ N, A and M lies in a straight line.

Hence, proved that M, A and N are collinear.

(ii) From equations (2) and (4), we get :

⇒ AM = AN.

Hence, proved that A is the mid-point of MN.

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a 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.

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,

E is the mid-point of BC and FE || AB.

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

Since,

⇒ FE || AB

∴ FE || BD.

D and F are mid-point of sides AB and AC respectively.

∴ DF || BC (By mid-point theorem)

∴ DF || BE.

Since, opposite sides of quadrilateral BEFD are parallel.

Hence, proved that BEFD is a parallelogram.

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that :

(i) triangles HEB and FHC are congruent;

(ii) GEHF is a parallelogram.

Answer

(i) In △ HEB and △ FHC,

⇒ BE = CF (Since, opposite sides of parallelogram are equal i.e. AB = CD and E and F are mid-points of AB and CD respectively)

⇒ ∠HBE = ∠HFC (Alternate angles are equal)

⇒ ∠EHB = ∠FHC (Vertically opposite angles are equal)

∴ △ HEB ≅ △ FHC (By A.A.S. axiom)

Hence, proved that triangles HEB and FHC are congruent.

(ii) Since,

△ HEB ≅ △ FHC

We know that,

Corresponding parts of congruent triangle are equal.

⇒ EH = CH and BH = FH.

⇒ H is the mid-point of BF and CE.

In △ AGE and △ DGF,

⇒ AE = DF (Since, opposite sides of parallelogram are equal i.e. AB = CD and E and F are mid-points of AB and CD respectively)

⇒ ∠GEA = ∠GDF (Alternate angles are equal)

⇒ ∠AGE = ∠DGF (Vertically opposite angles are equal)

∴ △ AGE ≅ △ DGF (By A.A.S. axiom)

∴ AG = GF and EG = DG [By C.P.C.T.C.]

⇒ G is the mid-point of DE and AF.

In △ ECD,

F and H are mid-points of sides CD and EC respectively.

∴ FH || DE [By mid-point theorem]

⇒ FH || GE

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

∴ GF || EC [By mid-point theorem]

⇒ GF || EH

Since, opposite sides of quadrilateral GEFH are parallel.

Hence, proved that GEHF is a parallelogram.

In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet side BC at points M and N respectively. Prove that : BM = MN = NC.

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.

By 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.

In △ AEG,

D is the mid-point of AE and DF || EG.

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

∴ AF = FG .........(1)

Since,

DF || EG || BC and DE || BE

∴ FG = GC [By equal intercept theorem]...........(2)

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

⇒ AF = FG = GC

Since, AB || FM || GN and AF = FG = GC

∴ BM = MN = NC [By equal intercept theorem]

Hence, proved that BM = MN = NC.

In triangle ABC; M is the mid-point of AB, N is mid-point of AC and D is any point in base BC. Use intercept theorem to show that MN bisects AD.

Answer

Let MN intersects at AD at point X.

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,

M is mid-point of AB and N is mid-point of AC.

∴ MN || BC (By mid-point theorem)

By 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.

Since,

M is mid-point of AB.

∴ AM = MB

N is mid-point of AC.

∴ AN = CN

From figure,

MN || BC, AM = BM and AN = CN

∴ AX = DX (By equal intercept theorem)

Hence, proved that MN bisects AD.

If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle, show that the diagonals AC and BD intersect at right angle.

Answer

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

Diagonal AC and BD intersect at point O.

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.

∴ PQ || AC (By mid-point theorem)

From figure,

⇒ ∠AOD = ∠PXO (Corresponding angles are equal) ............(1)

In △ BCD,

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

∴ QR || BD (By mid-point theorem)

Interior angles of a rectangle equals to 90°.

⇒ ∠RQX = ∠Q = 90°.

From figure,

⇒ ∠PXO = ∠RQX = 90° (Corresponding angles are equal) ............(2)

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

⇒ ∠AOD = ∠PXO = 90°.

Hence, AC and BD intersect at right angles.