Question

In the adjoining figure, ABCD is a parellelogram. E and F are mid points of the sides AB and CD respectively . The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively . Prove that (1) HEB is congruent to HCF (2) GEHF is a parellelogram. ​

Answer 1

Step-by-step explanation:

ANSWER

Given that,

ABCD is a parallelogram

E and F are the mid points of sides AB and CD

To prove:

△HEB≅△HCF

Proof:

in parallelogram ABCD,

∠CEB=∠ECF [ Alternate angles ]

⇒∠HEB=∠HCF...(1)

∠EBF=∠CFB [ Alternate angles ]

⇒∠EBH=∠CFM...(2)

Also, E and F are mid point of AB and CD

BE=

2

1

AB...(3)

CF=

2

1

CD...(4)

We know that

ABCD is a parallelogram

⇒AB=CD

⇒

2

1

AB=

2

1

CD

Using equation (3) and (4),

⇒BE=CF...(5)

Therefore In △HEB and △HCF,

∠HEB=∠FCH [ from eq (1)]

∠EBH=∠CFH [ from eq (2)]

BE=CF [ from eq (5)]

Hence, △HEB≅△FCH [ASA

Answer 2

Answer:

1) to prove ΔHEB ≅ ΔHCF

AB = DC (opposite sides of a parallelogram)

1/2 AB = 1/2 DC              EB = FC

AB║DC ; FC║EB ⇒∠HEB = ∠HCF (Alternate interior angles )

∠FHC = ∠EHB (Vertically opposite angles)

⇒ΔHEB ≅ ΔHCF