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Quadrilaterals

Long Answer Type

41. In a parallelogram ABCD, E and Fare the mid-points of sides AB and CD respectively. (see figure). Show that the line segments AF and EC trisect the diagonal BD.

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

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43.

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) $CM space equals space MA space equals space 1 half AB.$

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Short Answer Type

44. In triangle ABC, points M and N on sides AB and AC respectively are taken so that

AM equals 1 fourth AB space and space AN equals 1 fourth AC comma space

Prove that MN =

1 fourth

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45. In triangle ABC, points M and N on sides AB and AC respectively are taken so that

AM equals 1 fourth AB space and space AN equals 1 fourth AC comma space Prove space that space MN equals 1 fourth

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46. ABCD is a rhombus and AB is produced to E and F such that AE = AB = BF. Prove that ED and FC are perpendicular to each other.

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Long Answer Type

47. In ∆ABC, AD is the median through A and E is the mid-point of AD. BE is produced to meet AC in F. Prove that

AF equals 1 third AC

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48.

D, E, F are respectively the midpoints of the sides BC, CA and AB of a triangle ABC. Show that:
(i) BDEF is a parallelogram
(ii) DFEC is a parallelogram.

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Short Answer Type

49. In the figure ABCD is a parallelogram and E is the mid-point of side BC. DE and AB on producing meet at F. Prove that AF = 2AB.

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Long Answer Type

50. ABCD is a trapezium in which side AB is parallel to the side DC and E is the mid-point of side AD (see figure). If F is a point on the side BC such that the segment EF is parallel to the side DC, prove that F is the mid-point of BC and EF = $1 half$ (AB + DC).

Given: ABCD is a trapezium in which side AB is parallel to the side DC and E is the mid-point of side AD. F is a point on the side BC such that the segment EF is parallel to the side DC.

To Prove: F is the mid-point of BC

and    $EF equals 1 half left parenthesis AB plus DC right parenthesis$

Given: ABCD is a trapezium in which side AB is parallel to the side D

Construction: Join AC to intersect EF at G.
Proof:
(i) In ∆ADC,
∵ E is the mid-point of AD and
EG || DC    | ∵ EF || DC
∴ G is the mid-point of AC
| by converse of mid-point theorem Again, In ∆CAB,
∵ G is the mid-point of AC | proved above
and    GF || DC    | ∵ EF || DC
∴ F is the mid-point of BC | by converse of
mid-point theorem
(ii) We shall prove that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and half of it.

Given: ABCD is a trapezium in which side AB is parallel to the side D

Let ABC be a triangle in which D and E are the mid-points of the sides AB and AC respectively.
We produce DE to F such that DE = EF and join FC.
In ∆AED and ∆CEF,
AE = CE
| ∵ E is the mid-point of AC
∠AED = ∠CEF
| Vertically opposite angles
ED = EF    | by construction
∴ ∆AED ≅ ∆CEF
| SAS congruence rule
∴ AD = CF    |CPCT
and ∠ADE = ∠CFE    | CPCT
Now, ∵ D is the mid-point of AB
∴ AD = DB
But AD = CF    | Proved above
∴ DB = CF    ...(1)
Again, ∵ DF intersects AD and FC such that
∠ADE = ∠CFE    | Proved above
But these angles form a pair of equal alternate interior angles
∴ AD || FC
⇒ DB || FC    ...(2)
In view of (1) and (2),
DBCF is a parallelogram
| A quadrilateral is a parallelogram if one pair of its opposite sides is parallel and equal
∴ DF = BC
| Opposite sides of a parallelogram are equal
⇒ 2DE = BC    | by construction

$rightwards double arrow space space space DE equals 1 half BC$

Also, ∵ DBCF is a parallelogram
∴ DF || BC
| Opposite sides of a parallelogram are parallel
⇒ DE || BC
(iii) Using this result in ∆ADC,

$EG equals 1 half DC space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 3 right parenthesis$

| ∵ E is the mid-point of AD and G is the mid-point of AC
and in ∆CAB,

$GF equals 1 half AB space space space space space space space space space space space space space space space space space space space space space... left parenthesis 4 right parenthesis$

| ∵ G is the mid-point of AC and F is the mid-point of BC
Adding (3) and (4), we get

$EG plus GF equals 1 half DC plus 1 half AB rightwards double arrow space space space EF equals 1 half left parenthesis AB plus DC right parenthesis$