\(ABCD\) is a rectangle in which diagonal \(AC\) bisects \(∠A\) as well as \(∠C\).
Verify that:
(i) All the sides of \(ABCD\) are equal, and hence it is a square.
(ii) The other diagonal \(BD\) also acts as an angle bisector of \(\angle B\) and \(\angle D\).
Given: \(ABCD\) be a rectangle such that \(AC\) bisects \(∠A\) as well as \(∠C\).
So \(∠BAC = ∠DAC\) and, \(∠DCA = ∠BCA\) -----(1)
Explanation:
(i) To Prove: \(ABCD\) is a square.
Proof:
As we know that every rectangle is a parallelogram.
\(ABCD\) is a parallelogram.
\(∠BCA = \) -----(2) [ ]
From (1) and (2), we have
\(∠DCA= ∠\) -----(3)
In \(∆DCA\),
\(∠DCA= ∠DAC\) then, \(CD =\) [Sides opposite to equal angles of a triangle are equal]
Similarly, in \(∆ABC\), \(\angle ACB=\angle BAC\) then, \(AB =\)
So, \(ABCD\) is a rectangle having adjacent sides equal.
\(ABCD\) is a square.
(ii) To Prove: Diagonal \(BD\) bisects \(∠B\) as well as \(∠D\).
Proof:
Since, \(ABCD\) is a square
\(AB = BC = CD = DA\)
So, In \(∆ABD\), as \(AB = AD\)
\(∠ABD = ∠\) [Angles opposite to equal sides of a triangle are equal] -----.(1)
Similarly, in \(\bigtriangleup BCD\), \(∠CBD = ∠\) ------(2)
\(∠CBD = ∠ADB\) [] -----(3)
From (1) and (3)
\(∠CBD = ∠\)
From (2) and (3)
\(∠ADB = ∠\)
So, \(BD\) bisects \(∠B\) as well as \(∠D\).