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Question

ABCD is a rectangle in which diagonal AC bisects A as well asC. Show that:

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

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Solution

Step 1: Drawing the diagram:

ABCD is a rectangle in which diagonal AC bisects A as well asC.

Join diagonal BD.

Step 2: Proving ABCD is a square:

As, diagonal AC bisects A as well asC,

BAC=DACBCA=DCA................(1)

As we know that every rectangle is a parallelogram.

ABCD is a parallelogram.

BCA=DAC...................(2) [ Alternate interior angles are equal]

From (1) and (2), we have

DCA=DAC...................(3)

In ADC,

DCA=DAC then,

AD=CD [Sides opposite to equal angles of a triangle are equal]

Similarly, AB=BC

In, rectangle ABCD

AB=CD,AD=BC (opposite sides of rectangle are equal)

AB=BC=CD=DA

Therefore, ABCD is a square.

Step 3: Proving Diagonal BD bisects B as well as D:

In ADB,

AD=AB (Side of square)

ABD=ADB (Angles opposite to equal sides are equal)

In CDB,

CD=CB (Side of square)

CBD=CDB (Angles opposite to equal sides are equal)

And, ABD=CDB [ Alternate interior angles are equal]

ABD=CBDADB=CDB

Therefore, Diagonal BD bisects B as well as D.

Hence proved.


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