Question

prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle...... batao​

Answer 1

Quadrilateral ABCD is a parallelogram,

Adjacent angles are supplementry

Therefore,

<ADC +<BCD =180°

Multiply by 1/2,

1/2×ACD+1/2×BCD=1/2×180 {1}

But,

1/2<ADC=<PDC (angle bisector

1/2<BCD=<PCD theorem) {2}

From {1}&{2},

<PDC+<PCD=90°

In triangle PDC,

<PDC+<PCD+<DPC=180°°

90°+<DPC=180° {3}

<DPC=90°

i.e<P&<R=90° {4}

<BDA+<ADC=180°

Multiply by 1/2

1/2<BAD+1/2<ADC=90° {5}

1/2<BAD=<SDA,1/2<ADC=<SDA, {6}

From {5}&{6}

<SDA+<SAD=90°

In triangle Sad,

<SAD+<SDA+<ASD=180°

90°+<ASD=180°

<ASD=90°

Therefore, <PSR=90°

i.e<S=<Q=90° {8 }

From {4} & {8},

<Q=<R=<O=<S=90°

i.e, Quadrilateral PQRS is a rectangle

See fig 5.24

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Is this answer in 9th standard textbook of state board