English

D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E. Prove that BDFE is a parallelogram Find AB, if EF = 4.8 cm.

Advertisements
Advertisements

Question

D and F are midpoints of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.

  1. Prove that BDFE is a parallelogram
  2.  Find AB, if EF = 4.8 cm.
Sum
Advertisements

Solution

The required figure is shown below

(i) Since F is the midpoint and EF || AB.

Therefore E is the midpoint of BC.

So, `BE = 1/2BC and EF = 1/2AB`   …..(1)

Since D and F are the mid-points of AB and AC

Therefore DE || AC.

So, `DF = 1/2BC and DB = 1/2"AB"`  …..(2)

From (1), (2) we get

BE = DF and BD = EF

Hence  BDEF is a parallelogram.

(ii) Since

AB = 2EF

= 2 × 4.8

= 9.6 cm.

shaalaa.com
  Is there an error in this question or solution?
Chapter 12: Mid-point and Its Converse [ Including Intercept Theorem] - Exercise 12 (A) [Page 151]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]
Exercise 12 (A) | Q 14 | Page 151

RELATED QUESTIONS

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


ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.


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.


In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.


In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF.
Show that: EF = AC.


In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.


ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.


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: EGFH is a parallelogram.


In ΔABC, D and E are the midpoints of the sides AB and BC respectively. F is any point on the side AC. Also, EF is parallel to AB. Prove that BFED is a parallelogram.

Remark: Figure is incorrect in Question


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×