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

Dear Student!

Given: ABCD is a rectangle. Diagonal AC bisects ∠A and ∠C.

To prove:

(1) ABCD is a square.

(2) Diagonal BD bisects ∠B and ∠D.

Proof:

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))

In ΔADC,

∠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.

Cheers!

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