Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.
To Prove :
(i) ABCD is a square.
(ii) Diagonal BD bisects ∠B as
well as ∠D.
Proof : (i) In ∆ABC and ∆ADC, we have
∠BAC = ∠DAC [Given]
∠BCA = ∠DCA [Given]
AC = AC
∴ ∆ABC ≅ ∆ADC [ASA congruence]
∴ AB = AD and CB = CD [CPCT]
But, in a rectangle opposite sides are equal,
i.e., AB = DC and BC = AD
∴ AB = BC = CD = DA
Hence, ABCD is a square Proved.
(ii) In ∆ABD and ∆CDB, we have
AD = CD
AB = CD [Sides of a square]
BD = BD [Common]
∴ ∆ABD ≅ ∆CBD [SSS congruence]
So, ∠ABD = ∠CBD
∠ADB = ∠CDB
Hence, diagonal BD bisects ∠B as well as ∠D Proved
