ABCD is a rectangle in which diagonal AC bisects angle A as well as angle C.show that
i) ABCD is a square
i) Diagonal BD bisects angle B as well as angle D
Given: ABCD is a rectangle. Diagonal AC bisects ∠A and ∠C.
(1) ABCD is a square.
(2) Diagonal BD bisects ∠B and ∠D.
AC bisects ∠A,
∴ ∠BAC = ∠DAC ... (1)
∠BCA = ∠DCA ... (2)
AB || CD and AC is the transversal,
∴ ∠BAC = ∠DCA (Alternate interior angles)
⇒ ∠DAC = ∠DCA ( Using (1))
∠DAC = ∠DCA
∴ CD = AD ... (3) (In a triangle, equal angles have equal sides opposite to them)
AB = CD and BC = DA ... (4) (Opposite sides of rectangle are equal)
From (3) and (4), we get
AB = BC = CD = DA
∴ ABCD is a square.
In ΔBAD and ΔBCD,
AB = CD (Given)
AD = BC (Given)
BD = BD (Common)
∴ ΔBAD ΔBCD (SSS Congruence Criterion)
⇒ ∠ABD = ∠CBD (CPCT)
and ∠ADB = ∠CDB (CPCT)
Thus, the diagonal BD bisects ∠B and ∠D.