Question

ABCD is a rectangle in which diagonal AC bisects angle A as well as angle C show that ABCD is a square and diagonal BD bisects Angle B as well as angle d

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

Figure:-

Given:-

• ABCD is a rectangle.
• Diagonal AC bisect <A as well as <C.

To find:-

1. ABCD is a square
2. Diagonal BD bisects <B as well <D

Solutions:-

Let ABCD is a rectangle

.:. <A = <C

1/2 <A = 1/2 <C

<DAC = <DCA (AC bisects <A and <C)

CD = DA (side opposite to equal angles area also equal)

DA = BC and AB = CD (opposite sides of a rectangle are equal)

AB = BC = CD = DA

ABCD is a rectangle and all of its sides are equal.

Hence, ABCD is a square.

Let us BA join.

In ∆BCD,

BC = CD (sides of a square are equal to each other)

<CDB = <CBA (Angle opposite to equal sides are equal)

<CDB = <ABD (Alternate interior angle for AB//CD)

.:. <CBD = <ABD

BD bisects <B

Also,

<CBD = <ADB (Alternate interior angle for BC//AD)

<CDB = <ABD

.:. BD bisects <D