Mid-point Theorem — Chapter Test

ABCD is a rhombus with P, Q and R as mid-points of AB, BC and CD respectively. Prove that PQ ⊥ QR.

Answer

Join AC and BD.

Diagonals of rhombus intersect at right angle.

∠MON = 90°

In △BCD,

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

RQ || DB and RQ = $\dfrac{1}{2}$DB

RQ || DB ⇒ MQ || ON

From figure,

∠MON + ∠MQN = 180° (Sum of alternate angles of quadrilateral = 180°)

∠MQN = 180° - 90° = 90°

∴ PQ ⊥ QR.

Hence, proved that PQ ⊥ QR.

The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.

Answer

From figure,

ABCD is a quadrilateral in which diagonals AC and BD are perpendicular to each other. P, Q, R and S are mid-points of AB, BC, CD and DA.

In △ABC,

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

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

In △ADC,

S and R are mid-points of AD and DC respectively,

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

Using eqn. 1 and 2 we get,

PQ || SR and PQ = SR.

So, PQRS is a parallelogram.

In △ABD,

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

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

Given,

AC and BD intersect at right angles,

From 3 we get,

SP || BD.

∴ SP ⊥ AC

From 2 we get,

SR || AC

∴ SP ⊥ SR i.e. ∠RSP = 90°.

∴ PQRS is a rectangle.

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

If D, E and F are mid-points of the sides BC, CA and AB respectively of a △ABC, prove that AD and FE bisect each other.

Answer

△ABC with D, E and F as mid-points of the sides BC, CA and AB is shown below:

D and E are midpoints of BC and CA respectively,

DE = $\dfrac{1}{2}$AB and DE || AB or DE || AF .......(1)

Since,

F is midpoint of AB,

AF = $\dfrac{1}{2}$AB

∴ AF = DE .......(2)

F and D are midpoints of AB and BC respectively,

FD = $\dfrac{1}{2}$AC and FD || AC or FD || AE .......(3)

Since,

E is midpoint of AC,

AE = $\dfrac{1}{2}$AC

∴ FD = AE .......(4)

From 1, 2, 3 and 4 we get,

DE || AF, AF = DE and FD || AE, FD = AE.

Hence, AEDF is a parallelogram.

∴ AD and EF bisect each other.

Hence, AD and EF bisect each other.

In △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

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

DE || BC or DE || BF and DE = $\dfrac{1}{2}$BC ......(By midpoint theorem) .....(i)

Given, through E, a straight line is drawn parallel to AB to meet BC at F.

F will be mid-point of BC (By converse of mid-point theorem).

Since, F is midpoint of BC,

∴ BF = $\dfrac{1}{2}$BC .....(ii)

From (i) and (ii) we get,

DE = BF and DE || BF.

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

FE || AB or FE || BD and FE = $\dfrac{1}{2}$AB ......(By midpoint theorem) .....(iii)

Since, D is midpoint of AB,

∴ BD = $\dfrac{1}{2}$AB .....(iv)

From (iii) and (iv) we get,

BD = FE and BD || FE.

Since, DE = BF, DE || BF and BD = FE, BD || FE

Hence, proved that BDEF is a parallelogram.

Perimeter of BDEF = BD + DE + FE + BF = BD + DE + BD + FE = 2(BD + FE).

BD = $\dfrac{1}{2}$AB = $\dfrac{1}{2}(8)$ = 4 cm.

FE = $\dfrac{1}{2}$BC = $\dfrac{1}{2}(9)$ = 4.5 cm.

Perimeter of BDEF = 2(BD + FE) = 2(4 + 4.5) = 2 × 8.5 = 17 cm.

Hence, perimeter of BDEF = 17 cm.

In the adjoining figure, ABCD is a parallelogram and E is mid-point of AD. DL || EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.

Answer

Given, DL || EB.

Since, DL || BE we can say that,

⇒ BE || DF

In △AFD,

E is midpoint of AD and BE is parallel to DF,

∴ B is midpoint of AF (By converse of midpoint theorem).

In BEDL,

LD || BE and BL || DE

∴ BEDL is a parallelogram.

Since, BEDL is a parallelogram opposite sides are equal.

Let LD = BE = x.

E is midpoint of AD and B is the midpoint of AF

By midpoint theorem,

BE = $\dfrac{1}{2}$FD

FD = 2BE = 2x.

LF = FD - LD = 2x - x = x.

Since, LF = BE = x.

Hence, proved that B is midpoint of AF and EB = LF.

In the adjoining figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively. Show that CR = $\dfrac{1}{4}$AC.

Answer

In parallelogram, diagonals bisect each other.

∴ AO = OC = $\dfrac{1}{2}$AC ......(i)

In △BCD,

P and Q are midpoints of CD and BC,

PQ || BD (By midpoint theorem)

Since, PQ || BD

∴ QR || BO

In △BCO,

Q is midpoint of BC and QR || BO

∴ R is midpoint of OC

CR = $\dfrac{1}{2}$OC

Substituting value of OC from (i) in above equation,

CR = $\dfrac{1}{2}$OC = $\dfrac{1}{2} \times \dfrac{1}{2} \times$ AC = $\dfrac{1}{4}$AC.

Hence, proved that CR = $\dfrac{1}{4}$AC.