Rectangle ABCD is shown in the figure below:
(i) Given :
ABCD is a rectangle and AC bisects ∠A and ∠C.
⇒ ∠DAC = ∠CAB ....(1)
⇒ ∠DCA = ∠BCA .....(2)
We know that,
Opposite sides of a rectangle are parallel and equal.
From figure,
AD || BC and AC is transversal,
⇒ ∠DAC = ∠BCA ..........(3) (Alternate interior angles are equal)
From equations (1) and (3), we get :
⇒ ∠CAB = ∠BCA ......(4)
In △ ABC,
⇒ ∠CAB = ∠BCA
We know that,
Sides opposite to equal angles are equal.
⇒ BC = AB .....(5)
We know that,
Opposite sides of a rectangle are equal.
⇒ BC = AD .........(6)
⇒ AB = DC .........(7)
From equation (5), (6) and (7), we get :
⇒ AB = BC = CD = AD.
Since,
ABCD is a rectangle and all the sides are equal. Hence, ABCD is a square.
Hence, proved that ABCD is a square.
(ii) Join BD.
In Δ BCD,
⇒ BC = CD (Sides of a square are equal to each other)
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal) ..... (8)
⇒ ∠CDB = ∠ABD (Alternate interior angles are equal) ..... (9)
From equations (8) and (9), we get :
⇒ ∠CBD = ∠ABD
∴ BD bisects ∠B.
From figure,
⇒ ∠CBD = ∠ADB (Alternate interior angles are equal) .....(10)
From equations (8) and (10), we get :
⇒ ∠ADB = ∠CDB
∴ BD bisects ∠D.
Hence, proved that diagonal BD bisects ∠B as well as ∠D.