English

Prove that the Straight Lines Joining the Mid-points of the Opposite Sides of a Quadrilateral Bisect Each Other. - Mathematics

Advertisements
Advertisements

Question

Prove that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Sum
Advertisements

Solution


Join AC.

P and Q are mid-points of AB and BC respectively.

∴ PQ || AC, PQ = `(1)/(2)"AC"`.........(i)

S and R are mid-points of AD and DC respectively.

∴ SR || AC, SR = `(1)/(2)"AC"`.........(ii)
From (i) and (ii)
PQ = SR
Therefore, PQRS is a parallelogram.
Since, diagonals of a parallelogram bisect each other
Therefore, PQ and QS bisect each other.

shaalaa.com
  Is there an error in this question or solution?
Chapter 15: Mid-point and Intercept Theorems - Exercise 15.1

APPEARS IN

Frank Mathematics [English] Class 9 ICSE
Chapter 15 Mid-point and Intercept Theorems
Exercise 15.1 | Q 8

RELATED QUESTIONS

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 a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.


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 given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.

[Hint: DN || QM]


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.

ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.


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.


AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF: AC = 1 : 3.


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


The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×