मराठी

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. - Mathematics

Advertisements
Advertisements

प्रश्न

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.

बेरीज

उत्तर

Given: In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively.

Also, AC = BD

To prove: PQRS is a rhombus.


Proof: In ΔADC, S and R are the mid-points of AD and DC respectively.

Then, by mid-point theorem.

SR || AC and SR = `1/2` AC  ...(i)

In ΔABC, P and Q are the mid-points of AB and BC respectively.

Then, by mid-point theorem.

PQ || AC and PQ = `1/2` AC  ...(ii)

From equations (i) and (ii),

SR = PQ = `1/2` AC  ...(iii)

Similarly, in ΔBCD,

RQ || BD and RQ = `1/2` BD  ...(iv)

And in ΔBAD,

SP || BD and SP = `1/2` BD  ...(v)

From equations (iv) and (v),

SP = RQ = `1/2` BD = `1/2` AC  [Given, AC = BD] ...(vi)

From equations (iii) and (vi),

SR = PQ = SP = RQ

It shows that all sides of a quadrilateral PQRS are equal.

Hence, PQRS is a rhombus.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Quadrilaterals - Exercise 8.4 [पृष्ठ ८२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
पाठ 8 Quadrilaterals
Exercise 8.4 | Q 3. | पृष्ठ ८२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

  1. SR || AC and SR = `1/2AC`
  2. PQ = SR
  3. PQRS is a parallelogram.


In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.


BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.


In ΔABC, D is the mid-point of AB and E is the mid-point of BC.

Calculate:
(i) DE, if AC = 8.6 cm
(ii) ∠DEB, if ∠ACB = 72°


In ΔABC, AB = 12 cm and AC = 9 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC in N, what is the length of MN?


Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.


D, E and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC. Prove that ΔDEF is also isosceles.


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: F is the mid-point of BC.


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 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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×