Question

6. 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.​

Answer 1

Given:

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

To prove:

(i) Triangles HEB and FHC are congruent

(ii) GEHF is a parallelogram

Solution:

(i) Proving triangles HEB and FHC are congruent:

In parallelogram ABCD,

E is the midpoint of the side AB

∴ AE = BE

and

F is the midpoint of side CD  

∴ DF = CF

Also, AB = CD ..... [opposite sides of parallelogram are equal in length]

∴ AE = BE = DF = CF  ...... (1)  

In △HEB and △FHC,  we have

∠EHB = ∠FHC ....... [ vertically opposite angles ∵ AB // CD]

∠HFC = ∠HBE ..... [alternate angles ∵ AB // CD and BF is a transversal]

BE = CF  ..... [from (1)]

∴ Δ HEB ≅ Δ FHC ..... [by AAS congruency]

Hence proved

(ii). Proving the GEHF is a parallelogram:

Considering quadrilateral AECF,  we have

AE = CF  ..... [from (1)]

and

AE // CF ...... [AB // CD]

We can conclude,  

⇒ AECF is a parallelogram ..... [∵ Opposite sides of a parallelogram are parallel to each other and equal in length]

⇒ EC // AF

⇒ EH // GF ...... (2)

Similarly, considering quadrilateral BEDF,  we have

BE = DF  ..... [from (1)]

and  

BE // DF .... [AB // CD)]

We can conclude,  

⇒ BEDF is a parallelogram ..... [∵ Opposite sides of a parallelogram are parallel to each other and equal in length]

⇒ BF // ED

⇒ HF // GE ..... (3)

From (2) and (3) we get,

EH // GF and HF // GE

⇒ since the quadrilateral, GEHF has its pair of opposite sides parallel to each other and in a parallelogram, the pair of opposite sides are parallel to each other and equal in length.

∴ GEHF is a parallelogram

Hence proved

[Note:- Figure is given as an attachment below]

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