Mid-Point Theorem & Intercept Theorem — Competency Focused Questions

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a square only, if :

1. ABCD is a rhombus

2. Diagonals of ABCD are equal

3. Diagonals of ABCD are equal and perpendicular

4. Diagonals of ABCD are perpendicular

Answer

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 mid-points of AB and BC respectively.

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

Similarly in △ACD,

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

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

In △ABD,

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

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

Similarly in △BCD,

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

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

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

PQ = SR = SP = QR

∴ PQRS is a rhombus.

Since, SP || BD and AC ⊥ BD

∴ SP ⊥ AC

⇒ SN ⊥ AC

⇒ ∠SNO = 90°

Since, SR || AC and AC ⊥ BD

∴ SR ⊥ BD

⇒ SM ⊥ BD

⇒ ∠SMO = 90°

From figure,

⇒ ∠MOC + ∠MON = 180° [Linear pair]

⇒ 90° + ∠MON = 180°

⇒ ∠MON = 180° - 90° = 90°.

In quadrilateral sum of angles = 360°

⇒ ∠O + ∠M + ∠N + ∠S = 360°

⇒ 90° + 90° + 90° + ∠S = 360°

⇒ 270° + ∠S = 360°

⇒ ∠S = 360° - 270°

⇒ ∠S = 90°.

Since, in rhombus adjacent angles sum = 180°

Thus, in rhombus PQRS.

⇒ ∠S + ∠R = 180°

⇒ 90° + ∠R = 180°

⇒ ∠R = 180° - 90°

⇒ ∠R = 90°.

⇒ ∠Q + ∠R = 180°

⇒ 90° + ∠Q = 180°

⇒ ∠Q = 180° - 90°

⇒ ∠Q = 90°.

⇒ ∠S + ∠P = 180°

⇒ 90° + ∠P = 180°

⇒ ∠P = 180° - 90°

⇒ ∠P = 90°.

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

∴ PQRS is a square.

Thus, we can say that :

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.

Hence, option 3 is the correct option.

D and E are the mid-points of the sides AB and AC respectively of △ABC. DE is produced to F. To show that CF is equal and parallel to DA, we need an additional information, which is :

1. DE = EF

2. AE = EF

3. ∠DAE = ∠EFC

4. ∠ADE = ∠ECF

Answer

Assume that, DE = EF

In △ADE and △CFE,

⇒ AE = CE

⇒ DE = EF

⇒ ∠AED = ∠CEF (Vertically opposite angles are equal)

∴ △ADE ≅ △CFE (S.A.S. axiom)

⇒ DA = CF (Corresponding parts of congruent triangles are equal)

⇒ ∠DAE = ∠ECF ..(1) (Corresponding parts of congruent triangles are equal)

⇒ ∠ADE = ∠EFC ..(2) (Corresponding parts of congruent triangles are equal)

Since, ∠DAE and ∠ECF are alternate angles and since they are equal, thus DA // CF.

Thus, if DE = EF then DA is equal and parallel to CF.

Hence, option 1 is the correct option.

In which of the following cases you will get 2XY = QR for the given figure?

(i) When PX = QX and PY = RY

(ii) When PX = QX and a + b = 180°

1. Only in case (i)

2. Only in case (ii)

3. In both the cases

4. None of these

Answer

In case (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 △PQR,

Given,

PX = QX and PY = RY

⇒ X and Y are midpoints of PQ and PR respectively.

∴ XY || QR and XY = $\dfrac{1}{2}$ QR

⇒ QR = 2 XY

∴ Case (i) is true.

In case (ii) :

Given,

PX = QX

⇒ X is the mid-point of PQ

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,

⇒ ∠QXY + ∠XQR = 180°

⇒ a + b = 180°

⇒ ∠QXY and ∠XQR are co-interior angles and their sum is equal to 180°.

∴ XY is parallel to QR.

In △PQR,

Since, X is the mid-point of PQ and XY // QR, thus :

Y is mid-point of PR.

Since, X and Y are mid-points of side PQ and PR respectively.

⇒ XY = $\dfrac{1}{2}$ QR (By mid-point theorem)

⇒ QR = 2 XY

∴ Case (ii) is true.

Hence, option 3 is the correct option.

In the figure, R is the mid-point of AB, P is the mid-point of AR and L is the mid-point of AP. If RS, PQ and LM are parallel to each other, then the length of BC is :

1. 3 LM

2. 4 LM

3. 6 LM

4. 8 LM

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,

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

In △APQ,

Given,

AL = LP and LM || PQ

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

⇒ L and M are midpoints of AP and AQ respectively.

∴ LM = $\dfrac{1}{2}$ PQ

⇒ PQ = 2 LM ......(1)

In △ARS,

Given,

AP = RP and PQ || RS

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

⇒ P and Q are midpoints of AR and AS respectively.

∴ PQ = $\dfrac{1}{2}$ RS .........(2)

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

⇒ 2 LM = $\dfrac{1}{2}$ RS

⇒ RS = 4 LM ....(3)

In △ABC,

Given,

AR = BR

⇒ RS || BC and S is the mid-point of AC. (By converse of mid-point theorem)

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

Substituting value of RS in equation (3), we get:

⇒ $\dfrac{1}{2}$ BC = 4 LM

⇒ BC = 8 LM.

Hence, option 4 is the correct option.

In the given figure, △ABC is a scalene triangle in which ∠B = 90°. P is the mid-point of AB, PQ || BC and QM ⊥ BC. Which type of quadrilateral is PQMB?

Answer

Given,

∠B = 90° and PQ || BC

AB is the transversal.

⇒ ∠PBM = ∠APQ = 90° (Corresponding angles are equal)

⇒ ∠APQ + ∠BPQ = 180° (Linear pair)

⇒ 90° + ∠BPQ = 180°

⇒ ∠BPQ = 180° - 90°

⇒ ∠BPQ = 90°

Since, QM ⊥ BC

⇒ ∠QMB = 90°

In a quadrilateral PQMB,

⇒ ∠PBM + ∠BPQ + ∠QMB + ∠PQM = 360°

⇒ 90° + 90° + 90° + ∠PQM = 360°

⇒ 270° + ∠PQM = 360°

⇒ ∠PQM = 360° - 270°

⇒ ∠PQM = 90°.

All the angles of a quadrilateral = 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 AB and PQ || BC, thus :

Q is mid-point of AC.

In △ABC,

Since, Q is the mid-point of AC and QM || AB (as both are perpendicular to BC), thus :

Mi si mid-point of BC.

In △ABC,

Since, Q and M are mid-points of AC and BC respectively.

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

⇒ QM = PB ...(1)

In △ABC,

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

⇒ PQ = $\dfrac{1}{2}$ BC (By mid-point theorem)

⇒ PQ = BM ...(2)

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

Since, opposite sides are equal and all the interior angles equals to 90°.

∴ PQMB is a rectangle.

Hence, PQMB is a rectangle.

In the given figure, A, B, C and D are mid-points of PQ, QR, RS and PS respectively. E, F, G and H are mid-points of AB, BC, CD and AD respectively. Which type of quadrilaterals are ABCD and EFGH?

Answer

Join QS, PR, 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 △QRP,

A and B are midpoints of PQ and QR respectively.

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

Similarly in △PRS,

D and C are midpoints of PS and RS respectively.

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

In △PQS,

D and A are midpoints of PS and PQ respectively.

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

Similarly in △QRS,

B and C are midpoints of QR and SR respectively.

∴ BC || QS and BC = $\dfrac{1}{2}$ 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

Since, opposite sides are parallel and equal.

Thus, ABCD is a parallelogram.

In △ABC,

E and F are midpoints of AB and BC respectively.

∴ EF || AC and EF = $\dfrac{1}{2}$ AC (By midpoint theorem) .....(5)

Similarly in △ADC,

H and G are midpoints of AD and CD respectively.

∴ GH || AC and GH = $\dfrac{1}{2}$ AC (By midpoint theorem) .....(6)

In △ABD,

H and E are midpoints of AD and AB respectively.

∴ EH || BD and EH = $\dfrac{1}{2}$ BD (By midpoint theorem) .....(7)

Similarly in △BCD,

F and G are midpoints of BC and CD respectively.

∴ FG || BD and FG = $\dfrac{1}{2}$ BD (By midpoint theorem) .....(8)

From (5) and (6) we get,

EF = GH AND EF || GH

From (7) and (8) we get,

EH = FG and EH || FG

Since, opposite sides are parallel and equal.

Thus, EFGH is a parallelogram.

Hence, ABCD and EFGH are parallelograms.