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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 (sec figure), AC is a diagonal. Show that

(i)  $SR\parallel AC$    and SR = ${1 \over 2}AC$
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Sol. (i) Consider triangle ACD,
S and R are mid-points of sides AD and DC respectively.

$SR\parallel AC$ and SR = ${1 \over 2}AC$                            ….(i)

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]
(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴  $PQ\parallel AC$  and   PQ= ${1 \over 2}AC$                                  ….(ii)
[Reason same as above]

From (i) and (ii),

$SR\parallel AC$   and $PQ\parallel AC$     ⇒    $SR\parallel PQ$     …(iii)

and SR = ${1 \over 2}$ AC and PQ =${1 \over 2}$ AC   ⇒  SR = PQ.          …(iv)
(iii) $SR\parallel PQ$   and SR = PQ.                                                    [From (iii) and (iv)]

⇒  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.

Sol. First prove that PQRS is a parallelogram.
(i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴  $SR\parallel AC$    and    $SR = {1 \over 2}AC$
[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]
(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.
∴   $PQ\parallel AC$    and    $PQ = {1 \over 2}AC$        …(ii) [Reason same as above]
From (i) and (ii),
$SR\parallel PQ$   and  $PQ\parallel AC$     ⇒  $SR\parallel PQ$      …(iii)

and   $SR = {1 \over 2}AC$ and  $PQ = {1 \over 2}AC$   ⇒   SR = PQ.     …(iv)
(iii) $SR\parallel PQ$   and    SR = PQ.                                              [From (iii) and (iv)]

⇒  PQRS is a parallelogram.

PQRS is a parallelogram.

As  $PX\parallel YO$  and   $PY\parallel OX$ is a parallelogram.

⇒     ∠YPX = ∠YOX = 90°      [Diagonals of a rhombus bisect each other and are at right angles.]|

As in parallelogram PQRS,

∠SPQ    is 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.

Sol. Construction:
Join AC and BD.
As ABCD is a rectangle.

∴  AC = BD                                                              …(i)

Consider ΔABC, P and Q are midpoints of sides AB and BC respectively.

∴      $PQ\parallel AC$ and $PQ = {1 \over 2}AC$

Similarly, consider AADC, S and R are mid-points of sides AD and DC respectively.
∴   $SR\parallel AC$   and   $SR = {1 \over 2}AC$                              ….(iii)
From (ii) and (iii)

$PQ = SR = {1 \over 2}AC$                                                                  ……(iv)

Similarly, we can show

$PS = QR = {1 \over 2}BD$

From (i), (iv) and (v), we have PQ = QR .= RS = SP

∴   PQRS is a rhombus.

4. ABCD is a trapezium in which $AB\parallel DC$ , 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.

Sol. Consider  ΔADB, $AB\parallel EF$  ⇒  $AB\parallel EG$

⇒    G is mid-point of BD.                                                                                  …(i)

[ A line drawn through mid-point of one side, parallel to other bisects the third side.]

Consider triangle BCD,

$AB\parallel CD$   and   $EF\parallel AB$

⇒     $EF\parallel CD$  ⇒   $GF\parallel CD$

⇒      F is mid-point of BC.                                 [Reason same as above]

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.

Sol.  $AB = CD \Rightarrow {1 \over 2}AB = {1 \over 2}CD$

⇒  AE = CF

As AE = CF and   $AE = CF$                                                  [ $AB = CD$]

⇒   AECF is a parallelogram.

⇒   $AP\parallel CE$                                                                    …(i)

Consider triangle ABP,

E is mid-point of AB and $EQ\parallel AP$                       [From (i)]

⇒  Q is mid-point of BP [A line segment drawn through mid-point of one side of a triangle and parallel to other, bisects the third side.]

BQ = PQ                                                                                           …(ii)

Similarly, by considering triangle DCQ and proceeding as

above, we can show that

DP = PQ                                                                                            …(iii)

⇒     BQ = PQ = DP                                                                         [From (ii) and (iii)]

⇒     P and Q trisect BD.

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

Sol.
(i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴   $SR\parallel AC$    and SR $= {1 \over 2}$                       …(i)

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]
(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴  $PQ = AC$ and PQ $= {1 \over 2}AC$                                     [Reason same as above]
From (i) and (ii),

$SR\parallel AC$  and $PQ\parallel AC$  ⇒   $SR\parallel PQ$                      …(iii)

and $SR = {1 \over 2}AC$    and    $PQ = {1 \over 2}AC$  ⇒   SR = PQ               . …(iv)
(iii) $SR\parallel PQ$  and SR = PQ.                                    [From (iii) and (iv)]

⇒  PQRS is a parallelogram.

We know that diagonals of a parallelogram bisect each other, i.e., OP = OR and OQ = OS.
Hence, line segments joining midpoints of opposite sides of a quadrilateral bisect each other.

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 = ${1 \over 2}$  AB.

Sol.
(i) $MD\parallel BC$    meets AC at D.

∴  D is mid-point of AC.

[A line through the mid-point of a ID side of a triangle parallel to other bisects and third side.]

(ii) $MD\parallel BC$  and AC is transversal.

∴   ∠ADM =  ∠ACB.                           [ Corresponding angles ]

⇒   ∠ADM  =  90°                                 [ ∠ACB = 90° ]

⇒    MD  ⊥  AC.

(iii) Consider triangles ADM and CDM,
AD = DC                                                               [ From result (i) ]

MD is common.

∠ADM = ∠CDM                                   [90° each]                   [From result (ii)]

∴   MA = CM = ${1 \over 2}AB$     [ M is mid-point of AB]