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Question

Show that the bisector of angles of a parallelogram from a rectangle.

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Solution

Let ABCD is a parallelogram

In figure,

AR,BR,CP and DP are the bisectors of angles A,B,C and D respectively.

The bisectors AR,BR,CP and DP formed a quadrilateral PQRS.

To prove: PQRS is a parallelogram.

Proof:

we know that,
In parallelogram, sum of the consecutive angles is supplementary.

A+B=180

A+B2=1802

A+B2=90

A2+B2=90

Now, In ΔARB,

A2+B2+ARB=180 [Since, angle sum property of a triangle]

90+ARB=180

ARB=18090

ARB=90 ---(1)

Similarly,

C+D=180

C+D2=90

C2+D2=90

In ΔCPD,

C2+D2+CPD=180

90+CPD=180

CPD=18090

CPD=90 ---(2)

Now,

A+D=180

A+D2=90

A2+D2=90

In ΔASD

A2+D2+ASD=180

90+ASD=180

ASD=18090

ASD=90

RSP=ASD=90 [Since, vertically opposite angles] ---(3)

In Quadrilateral PQRS,

P+S+R+Q=360

90+90+90+Q=360

270+Q=360

Q=360270

Q=90 ---(4)

From (1), (2), (3) and (4),

In Quadrilateral PQRS,

P=Q=R=S=90

PQRS is a rectangle.

Hence, proved.


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