ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that (i) ABCD is a square, (ii) diagonal BD bisects ∠B as well as ∠D.
Given: In rectangle ABCD, AC bisects ∠A, i.e. ∠1 = ∠2 and AC bisects ∠C, i.e. ∠3 = ∠4.
To prove:
(i) ABCD is a square,
(ii) diagonal BD bisects ∠B as well as ∠D.
Proof:
(i)
Since, $A D \| B C$ (Opposite sides of a rectangle are parallel.)
So, $\angle 1=\angle 4$ (Alternate interior angles)
But, $\angle 1=\angle 2$ (Given)
So, $\angle 2=\angle 4$
In $\Delta A B C$,
Since, $\angle 2=\angle 4$
So, $B C=A B$ (Sides opposite to equal angles are equal.)
But these are adjacent sides of the rectangle ABCD.
Hence, ABCD is a square.
(ii)
Since, the diagonals of a square bisects its angles.
So, diagonals BD bisects ∠B as well as ∠D.
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