मराठी

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 - Mathematics

Advertisements
Advertisements

प्रश्न

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.

बेरीज

उत्तर

Given that AD = BC                                                    …..(1)

From the figure,
For triangle ADC and triangle ABD

2GH = AD and 2EF = AD, therefore 2GH = 2EF = AD …..(2)

For triangle BCD and triangle ABC

2GF = BC and 2EH=BC, therefore 2GF= 2EH = BC      …..(3)

From (1), (2) ,(3) we get,
2GH = 2EF = 2GF = 2EH
GH = EF = GF = EH
Therefore EFGH is a rhombus.
Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Mid-point and Its Converse [ Including Intercept Theorem] - Exercise 12 (A) [पृष्ठ १५०]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 12 Mid-point and Its Converse [ Including Intercept Theorem]
Exercise 12 (A) | Q 8 | पृष्ठ १५०

संबंधित प्रश्‍न

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.


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]


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.


The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:

PR = `[1]/[2]` ( AB + CD)


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.


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 triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.


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


ΔABC is an isosceles triangle with AB = AC. D, E and F are the mid-points of BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.


In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×