Question

ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that

(i) ABCD is a square,

(ii) Diagonal BD bisects ∠ B as well as ∠ D.

Answer 1

(i) We know that ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C.

So we get

∠ BAC = ∠ DAC …….. (1)

∠ BCA = ∠ DCA …….. (2)

We know that every rectangle is a parallelogram

So we get AB || DC and AC is a transversal

From the figure we know that ∠ BAC and ∠ DCA are alternate angles

∠ BAC = ∠ DCA

By considering equation (1)

We get

∠ DAC = ∠ DCA

Consider △ ADC

We know that the opposite sides of equal angles are equal

AD = CD

Since ABCD is a rectangle

We get AB = BC and CD = AD

So we get AB = BC = CD = AD

Therefore, it is proved that ABCD is a square.

(ii) Consider △ BAD and △ BCD

We know that AB = CD and AD = BC

BD is common i.e. BD = BD

By SSS congruence criterion

△ BAD ≅ △ BCD

We know that

∠ ABD = ∠ CBD and ∠ ADB = ∠ CDB (c. p. c. t)

Therefore, diagonal BD bisects ∠ B as well as ∠ D.