In the adjoining figure, ABCD is a parallelogram. 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

(i) △HEB ≅ △HCF

(ii) GEHF is a parallelogram.

(i) We know that,

ABCD is a parallelogram,

∴ FC || BE

∠CEB = ∠FCE (Alternate angles)

⇒ ∠HEB = ∠FCH …….(1)

∠EBF = ∠CFB (Alternate angles)

⇒ ∠EBH = ∠CFH …….(2)

Here E and F are mid-points of AB and CD

BE = $\dfrac{1}{2}$AB ……..(3)

CF = $\dfrac{1}{2}$CD ……..(4)

We know that ABCD is a parallelogram,

AB = CD

Now dividing both sides by $\dfrac{1}{2}$

$\dfrac{1}{2}$AB = $\dfrac{1}{2}$CD

Using equations 3 and 4 we get,

BE = CF …….(5)

In △HEB and △HCF,

∠HEB = ∠FCH (Using eqn. i)

∠EBH = ∠CFH (Using eqn. ii)

BE = CF (Using eqn. v)

∴ △HEB ≅ △HCF (By ASA axiom of congruency)

Hence, proved that △HEB ≅ △HCF.

(ii) AB = CD (As ABCD is a parallelogram)

Hence, AE = CF (As E and F are mid-points of the sides AB and CD respectively)

As, AB || CF we can say that,

AE || CF

Since, AE = CF and AE || CF

∴ AECF is a parallelogram.

∴ AF || EC

From figure we get,

GF || EH ……..(1)

AB = CD (As ABCD is a parallelogram)

Hence, DF = EB (As E and F are mid-points of the sides AB and CD respectively) and DF || EB.

Since, DF = EB and DF || EB

∴ DEBF is a parallelogram.

∴ DE || FB

From figure we get,

GE || FH ……..(2)

From 1 and 2 we get,

GF || EH and GE || FH.

∴ GEHF is a parallelogram.

Hence, proved that GEHF is a parallelogram.