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

Answer:

(i)ABCD is a rectangle , in which diagonal AC bisect ∠A as well as ∠C. Therefore,

∠DAC=∠CAB→(1)

∠DCA=∠BCA→(2)

A square is a rectangle when all sides are equal. Now,

AD∥BC & AC is transversal, therefore

∠DAC=∠BCA [Alternate angles]

From (1), ∠CAB=∠BCA→(3)

In △ABC,

∠CAB=∠BCA , therefore

BC=AB →(4)[sides opposite to equal angles]

But BC=AD & AB=DC→(5) [Opposite sides of rectangle]

Therefore from (4)& (5),

AB=BC=CD=AD

Hence, ABCD is a square.

(ii) ABCD is a square and we know that diagonals of a square bisect its

angles.

Hence, BD bisects ∠B as well as ∠D.

Answer 2

GIVEN : ABCD is a rectangle in which diagonal Ac bisects ∠ A as well as ∠ C. in rectangle ABCD AD=BC, AB=CD and ∠ A =∠ B = ∠ C= ∠ D = 90°

TO PROVE : (1) ABCD is a square.

(2) Diagonal BD bisects ∠ B as well as ∠ D

PROOF : (1) AB = BC and AB = CD

∠ 1 = ∠ 2 and ∠ 3 = ∠ 4 –––––——— equation 1

∠ 1 = ∠ 4 and ∠ 2 = ∠ 4 ( alternate interior angles ) —————————equation 2

From equation 1 and 2 we get

∠ 1 = ∠ 2 = ∠ 3 = ∠ 4

In ∆ ABC

∠ 2 = ∠ 4

so, AB= BC

In ∆ ACD

∠ 1 = ∠ 3

so, AD = CD

SO, AB=BC=CD=AD and ABCD is a square because all sides of rectangle are equal and hence it is a square.

(2) In ∆ ABD

AB=AD

So, ∠ 5=∠ 8 ----------------- equation3

In ∆ BCD

CD=BC

So, ∠ 6= ∠ 7------------------ equation 4

∠ 5=∠ 7 and ∠ 6 = ∠ 8 ( Alternate interior angles )

from equation 3,4 and 5 we get

∠ 5 = ∠ 6 = ∠ 7 = ∠ 8

so, ∠ 5=∠ 6

and ∠ 7 = ∠ 8

therefore, Diagonal BD bisects ∠ B as well as ∠D

Hence verified.