Exercise 8.2 Page: 113
1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.20). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
(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 = ½ 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 = ½ AC
also, SR = ½ AC
, PQ = SR
(iii) SR || AC ———————- from question (i)
and, PQ || AC ———————- from question (ii)
⇒ SR || PQ – from (i) and (ii)
also, PQ = SR
, PQRS is a parallelogram.
2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
Given in the question,
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.
To Prove,
PQRS is a rectangle.
Construction,
Join AC and BD.
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)
, ΔDRS ≅ ΔBPQ [SAS congruency]
RS = PQ [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)
, ΔQCR ≅ ΔSAP [SAS congruency]
RQ = SP [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
, PQRS is a parallelogram.
also, ∠PQR = 90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
, PQRS is a rectangle.
3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
Given in the question,
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.
Construction,
Join AC and BD.
To Prove,
PQRS is a rhombus.
Proof:
In ΔABC
P and Q are the mid-points of AB and BC, respectively
, PQ || AC and PQ = ½ AC (Midpoint theorem) — (i)
In ΔADC,
SR || AC and SR = ½ AC (Midpoint 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.
, QR || BD and QR = ½ BD (Midpoint 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.
Hence Proved
4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that F is the mid-point of BC.
Solution:
Given that,
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point 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 mid-points of sides AB and CD, respectively (see Fig. 8.22). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
Given that,
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD, respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,
ABCD is a parallelogram
, AB || CD
also, AE || FC
Now,
AB = CD (Opposite sides of parallelogram ABCD)
⇒½ AB = ½ 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 mid-point of DQ (Converse of mid-point theorem)
⇒ DP = PQ — (i)
Similarly,
In ΔAPB,
E is midpoint of side AB and EQ || AP (as AF || EC).
Q is the mid-point of PB (Converse of mid-point theorem)
⇒ PQ = QB — (ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
Hence Proved.
6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
Solution:
(i) In ΔACB,
M is the midpoint of AB and MD || BC
, D is the midpoint of AC (Converse of mid point theorem)
(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB = 90°
, ∠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)
, ΔAMD ≅ ΔCMD [SAS congruency]
AM = CM [CPCT]
also, AM = ½ AB (M is midpoint of AB)
Hence, CM = MA = ½ AB