﻿ Class 9 NCERT Math Solution
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TOPICS
Exercise - 8.2

Question-1 :-  ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Figure). AC is a diagonal. Show that :
(i) SR || AC and SR = AC/2
(ii) PQ = SR
(iii) PQRS is a parallelogram Solution :-
```(i) Given that: In ∆ADC, S and R are the mid-points of sides AD and CD respectively.
Proove that : SR || AC and SR = AC/2
Proof : In a triangle, the line segment joining the mid-points of any
two sides of the triangle is parallel to the third side and is half of it.
So, SR || AC and SR = AC/2.  .......(1)

(ii) Given that: In ∆ABC, P and Q are mid-points of sides AB and BC respectively.
Proove that : PQ = SR
Proof : Therefore, by using mid-point theorem,
PQ || AC and PQ = AC ... (2)
Using equations (1) and (2), we obtain
PQ || SR and PQ = SR ... (3)
So, PQ = SR

(iii) Proove that : PQRS is a parallelogram
Proof : From equation (3), we obtained PQ || SR and PQ = SR
Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.
Hence, PQRS is a parallelogram.
```

Question-2 :-  ABCD is a rhombus and P, Q, R and S are ©wthe mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution :-
``` Given that : In ∆ABC, P and Q are the mid-points of sides AB and BC respectively.
Prove that : Quadrilateral PQRS is a rectangle.
Proof : In ∆ABC,
R and S are the mid-points of CD and AD respectively.
PQ || AC and PQ = AC/2 (Using mid-point theorem......(1)
RS || AC and RS = AC/2 (Using mid-point theorem).....(2)
From equations (1) and (2), we obtain
PQ || RS and PQ = RS
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O.

MQ || ON ( PQ || AC)
QN || OM ( QR || BD)
Therefore, OMQN is a parallelogram.
∠MQN = ∠NOM
∠PQR = ∠NOM
However, NOM = 90° (Diagonals of a rhombus are perpendicular to each other)
∠PQR = 90°
Clearly, PQRS is a parallelogram having one of its interior angles as 90°.
Hence, PQRS is a rectangle.
```

Question-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 that : In ∆ABC, P and Q are the mid-points of AB and BC respectively.
Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal.
Construction : Let us join AC and BD.
Proof : In ∆ABC,
PQ || AC and PQ = AC/2 (Mid-Point Theorem).....(1)
SR || AC and SR =  AC (Mid-point theorem) ... (2)
Clearly, PQ || SR and PQ = SR
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other,
It is a parallelogram.
PS || QR and PS = QR (Opposite sides of parallelogram)... (3)
In ∆BCD, Q and R are the mid-points of side BC and CD respectively.
QR || BD and QR = BD (Mid-point theorem) ... (4)
However, the diagonals of a rectangle are equal.
AC = BD .........(5)
By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.
```

Question-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 Figure). Show that F is the mid-point of BC. Solution :-
``` Given that : Let EF intersect DB at G.
By converse of mid-point theorem, we know that a line drawn through the mid-point
of any side of a triangle and parallel to another side, bisects the third side.
Prove that : F is the mid-point of BC.
Proof : In ∆ABD,
EF || AB and E is the mid-point of AD.
Therefore, G will be the mid-point of DB.
As EF || AB and AB || CD,
EF || CD (Two lines parallel to the same line are parallel to each other)
In ∆BCD, GF || CD and G is the mid-point of line BD.
Therefore, by using converse of mid-point theorem, F is the mid-point of BC.
```

Question-5 :-  In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Figure). Show that the line segments AF and EC trisect the diagonal BD. Solution :-
```  Given that : ABCD is a parallelogram.
AB || CD
And hence, AE || FC
Again, AB = CD (Opposite sides of parallelogram ABCD)
AB/2 = CD/2
AE = FC (E and F are mid-points of side AB and CD)
Prove that : The line segments AF and EC trisect the diagonal BD.
Proof : In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other.
Therefore, AECF is a parallelogram.
AF || EC (Opposite sides of a parallelogram)
In ∆DQC, F is the mid-point of side DC and FP || CQ (as AF || EC).
Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
DP = PQ ... (1)
Similarly, in ∆APB, E is the mid-point of side AB and EQ || AP (as AF || EC).
Therefore, by using the converse of mid-point theorem, it can be said that
Q is the mid-point of PB.
PQ = QB ... (2)
From equations (1) and (2),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
```

Question-6 :- Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution :-
``` Given that : Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively.
Prove that : PR and QS bisect each other.
Construction : Join PQ, QR, RS, SP, and BD.
Proof : In ∆ABD, S and P are the mid-points of AD and AB respectively.
Therefore, by using mid-point theorem, it can be said that
SP || BD and SP = BD/2 ... (1)
Similarly  in  ∆BCD,
QR || BD and QR = BD/2 ... (2)
From equations (1) and (2), we obtain
SP || QR and SP = QR
In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other.
Therefore, SPQR is a parallelogram.
We know that diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.
```

Question-7 :- 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 = AB/2

Solution :-
``` Given that : ABCD is a rhombus.
Prove that : (i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = AB/2
Proof : (i) In ∆ABC,
It is given that M is the mid-point of AB and MD || BC.
Therefore, D is the mid-point of AC. (Converse of mid-point theorem)

(ii) As DM || CB and AC is a transversal line for them, therefore,
∠MDC + ∠DCB = 180° (Co-interior angles)
∠MDC + 90° = 180°
∠MDC = 90°
MD ⊥ AC (iii) Join MC.
In ∆AMD and ∆CMD,
AD = CD (D is the mid-point of side AC)