हिंदी

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: δGea ≅ - Mathematics

Advertisements
Advertisements

प्रश्न

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: ΔGEA ≅ ΔGFD

योग
Advertisements

उत्तर


Since ABCD is a parallelogram,
AB = CD and AD = BC
Now, E and F are the mid-points of AB and CD respectively,
⇒ AE = EB = DF = FC     ....(i)

In ΔGEA and ΔGFD,
AE = DF              ....[From (i)]
∠AGE = ∠DGF  ....(vertically opposite angles)
∠GAE = ∠GFD  ....(Alternate interior angles)
∴ ΔGEA ≅ ΔGFD.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Mid-point and Intercept Theorems - Exercise 15.2

APPEARS IN

फ्रैंक Mathematics [English] Class 9 ICSE
अध्याय 15 Mid-point and Intercept Theorems
Exercise 15.2 | Q 1.1

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.


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 triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of

the triangle formed by joining the mid-points of the sides of this triangle. 


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


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 trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.


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 the given figure, T is the midpoint of QR. Side PR of ΔPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR = `(1)/(4)"PQ"`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×