मराठी

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).

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.

बेरीज
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.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Quadrilaterals - EXERCISE 8.2 [पृष्ठ ११४]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 9
पाठ 8 Quadrilaterals
EXERCISE 8.2 | Q 4. | पृष्ठ ११४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.


In Fig. below, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.


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


Fill in the blank to make the following statement correct:

The triangle formed by joining the mid-points of the sides of a right triangle is            


If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle,
show that the diagonals AC and BD intersect at the right angle.


In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that

(i) R is the mid-point of BC, and

(ii) PR = `(1)/(2)"DB"`.


In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: AP = 2AR


In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: BC = 4QR


In ΔABC, X is the mid-point of AB, and Y is the mid-point of AC. BY and CX are produced and meet the straight line through A parallel to BC at P and Q respectively. Prove AP = AQ.


Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×