ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that :

(i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC ⊥ BD

Parallelogram ABCD is shown in the figure below:

(i) Given,

AC bisects ∠A,

∴ ∠CAB = ∠CAD = x (let) …….(i)

AB || CD (As opposite sides of parallelogram are parallel)

⇒ ∠DCA = ∠CAB (Alternate angles are equal)

∴ ∠DCA = x …….(ii)

and,

⇒ ∠BCA = ∠CAD (Alternate angles are equal)

∴ ∠BCA = x …….(iii)

From (i), (ii) and (iii) we get,

∠CAB = ∠CAD = ∠DCA = ∠BCA …….(iv)

Thus, ∠DCA = ∠BCA

Hence, proved that AC bisects ∠C.

(ii) From equation (iv) we get,

∠CAD = ∠DCA

∴ In △ADC,

DA = DC (Sides opposite to equal angles are equal)

However, DA = BC and AB = CD (opposite sides of a parallelogram are equal)

Thus, AB = BC = CD = DA

As ABCD is a parallelogram in which all sides are equal, therefore ABCD is a rhombus.

Hence, proved that ABCD is a rhombus.

(iii) In △OAB and △OCB,

OA = OC (Diagonals of a parallelogram bisect each other)

OB = OB (Common Side)

AB = BC (Sides of Rhombus)

∴ △OAB ≅ △OCB (SSS rule of congruency)

∴ ∠AOB = ∠BOC (c.p.c.t)

But ∠AOB + ∠BOC = 180°

⇒ ∠AOB + ∠AOB = 180°

⇒ ∠AOB = $\dfrac{180°}{2}$

⇒ ∠AOB = 90°

∴ AC ⊥ BD

Hence, proved that AC ⊥ BD.