Exercise 8.2
1. ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that :
(i) SR  AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Answer
(i) In ΔDAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SR  AC and SR = 1/2 AC
(ii) In ΔBAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQ  AC and PQ = 1/2 AC
also, SR = 1/2 AC
Thus, PQ = SR
(iii) SR  AC  from (i)
and, PQ  AC  from (ii)
⇒ SR  PQ  from (i) and (ii)
also, PQ = SR
Thus, PQRS is a parallelogram.
2. ABCD is a rhombus and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Answer
Given,
ABCD is a rhombus and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively.
To Prove,
PQRS is a rectangle.
Construction,
AC and BD are joined.
Proof,
In ΔDRS and ΔBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.
RS = PQ by CPCT  (i)
In ΔQCR and ΔSAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.
RQ = SP by CPCT  (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC respectively.
⇒ QR  BD
also,
P and S are the mid points of AD and AB respectively.
⇒ PS  BD
⇒ QR  PS
Thus, PQRS is a parallelogram.
also, ∠PQR = 90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
Thus, PQRS is a rectangle.
3. ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer
Given,
ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively.
Construction,
AC and BD are joined.
To Prove,
PQRS is a rhombus.
Proof,
In ΔABC
P and Q are the midpoints of AB and BC respectively
Thus, PQ  AC and PQ = 1/2 AC (Mid point theorem)  (i)
In ΔADC,
SR  AC and SR = 1/2 AC (Mid point theorem)  (ii)
So, PQ  SR and PQ = SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PS  QR and PS = QR (Opposite sides of parallelogram)  (iii)
Now,
In ΔBCD,
Q and R are mid points of side BC and CD respectively.
Thus, QR  BD and QR = 1/2 BD (Mid point theorem)  (iv)
AC = BD (Diagonals of a rectangle are equal)  (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.
4. ABCD is a trapezium in which AB  DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the midpoint of BC.
Answer
Given,
ABCD is a trapezium in which AB  DC, BD is a diagonal and E is the midpoint of AD.
To prove,
F is the midpoint of BC.
Proof,
BD intersected EF at G.
In ΔBAD,
E is the mid point of AD and also EG  AB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In ΔBDC,
G is the mid point of BD and also GF  AB  DC.
Thus, F is the mid point of BC (Converse of mid point theorem)
5. In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.
Answer
Given,
ABCD is a parallelogram. E and F are the midpoints of sides AB and CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,
ABCD is a parallelogram
Therefor, AB  CD
also, AE  FC
Now,
AB = CD (Opposite sides of parallelogram ABCD)
⇒ 1/2 AB = 1/2 CD
⇒ AE = FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AF  EC (Opposite sides of a parallelogram)
Now,
In ΔDQC,
F is mid point of side DC and FP  CQ (as AF  EC).
P is the midpoint of DQ (Converse of midpoint theorem)
⇒ DP = PQ  (i)
Similarly,
In APB,
E is mid point of side AB and EQ  AP (as AF  EC).
Q is the midpoint of PB (Converse of midpoint theorem)
⇒ PQ = QB  (ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
6. Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Answer
Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.
Now,
In ΔACD,
R and S are the mid points of CD and DA respectively.
Thus, SR  AC.
Similarly we can show that,
PQ  AC
PS  BD
QR  BD
Thus, PQRS is parallelogram.
PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.
7. ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the midpoint of AC
(ii) MD ⊥ AC
(iii) CM = MA = 1/2 AB
Answer
(i) In ΔACB,
M is the mid point of AB and MD  BC
Thus, D is the mid point of AC (Converse of mid point theorem)
(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB = 90°
Thus, ∠ADM = 90° and MD ⊥ AC
(iii) In ΔAMD and ΔCMD,
AD = CD (D is the midpoint of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (common)
Thus, ΔAMD ≅ ΔCMD by SAS congruence condition.
AM = CM by CPCT
also, AM = 1/2 AB (M is mid point of AB)
Hence, CM = MA = 1/2 AB
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