Advertisements
Advertisements
प्रश्न
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 the given figure). Show that F is the mid-point of BC.

Advertisements
उत्तर
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.
In ΔABD,
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 ΔBCD, 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.
APPEARS IN
संबंधित प्रश्न
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC =
21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.
The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)

In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.
The side AC of a triangle ABC is produced to point E so that CE = AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segment DL and BM trisect diagonal AC.
Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 3DF = EF
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: The line drawn through G and parallel to FE and bisects DA.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔHEB ≅ ΔHFC
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
