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.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

In △ ADC,

G and H are mid-points of sides CD and AC respectively.

⇒ GH = $\dfrac{1}{2}AD$

⇒ AD = 2GH ………(1)

In △ ABD,

E and F are mid-points of sides AB and BD respectively.

⇒ EF = $\dfrac{1}{2}AD$

⇒ AD = 2EF ………(2)

From equations (1) and (2), we get :

⇒ AD = 2GH = 2EF ……..(3)

In △ BCD,

G and F are mid-points of sides CD and BD respectively.

⇒ GF = $\dfrac{1}{2}BC$

⇒ BC = 2GF ………(4)

In △ ABC,

E and H are mid-points of sides AB and AC respectively.

⇒ EH = $\dfrac{1}{2}BC$

⇒ BC = 2EH ………(5)

From equations (4) and (5), we get :

⇒ BC = 2GF = 2EH ……..(6)

Given,

⇒ AD = BC ………….(7)

From equations (3), (6) and (7), we get :

⇒ 2GH = 2EF = 2GF = 2EH

⇒ GH = EF = GF = EH.

Since, all sides of quadrilateral EFGH are equal.

∴ EFGH is a rhombus.

Hence, proved that EFGH is a rhombus.