English

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.

Advertisements
Advertisements

Question

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.

Sum
Advertisements

Solution

The figure is shown below

AD = DB

BE = EC

EF || BD

In Δ ABC

E is the midpoint of AB and 

EF || BD

∴ By the midpoint theorem, F will be the midpoint of AC and D will be the midpoint of AB.

As D and F are midpoints of AC and AB respectively.

∴ By the midpoint theorem of DF || BC or BE

Since DF || BE and EF || BD

Hence, BEFD is a parallelogram.

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Mid-point Theorem and Its Converse [Including Intercept Theorem] - Exercise 12 (B) [Page 154]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 11 Mid-point Theorem and Its Converse [Including Intercept Theorem]
Exercise 12 (B) | Q 5 | Page 154

RELATED QUESTIONS

In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.


ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.


The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find: 

  1. MN, if AB = 11 cm and DC = 8 cm.
  2. AB, if DC = 20 cm and MN = 27 cm.
  3. DC, if MN = 15 cm and AB = 23 cm.

L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.


In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.


In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.


In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.


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: ∠EFG = 90°


In ΔABC, D, E and F are the midpoints of AB, BC and AC.
If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.


In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:

RT = `(1)/(3)"PQ"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×