ABCD is a rectangle in which diagonal BD bisects ∠B. Show that ABCD is a square.

Given, in a rectangle ABCD, diagonal BDl bisects ∠B.
Now, join AC.
In ΔBAD and ΔBCD,
∠ABD = ∠CBD
∠A = ∠C = 90![]()
And BD = BD (common)![]()
BAD
BCD
AB = BC
And AD = CD
But in rectangle opposite sides are equal,
AB = CD
And BC = AD
AB = BC = CD = DA
So, ABCD is a square.
Hence, Proved.