P,Q,R and S are respectively the mid-points of the sides AB ,BC, CD and DA of a quarilateral ABCD in which AC= BD, prove that PQRS is a rhombus.

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Solution

if $A C = B D$
then $A B C D$ must be a square or a rectangle
if $A B C D$ is a rectangle,
$P B = \frac{x}{2}, Q B = \frac{y}{2}$
$∴ P Q = \sqrt{P B^{2} + Q B^{2}}$ (Pythagorus theorem)
$= \sqrt{\frac{x^{2}}{4} + \frac{y^{2}}{4}}$
$P Q = \frac{1}{2} \sqrt{x^{2} + y^{2}}$
Similarly,
$A P = \frac{x}{2}, A S = \frac{y}{2}$
$P S = \sqrt{A P^{2} + A S^{2}} = \sqrt{\frac{x^{2}}{4} + \frac{y^{2}}{4}} = \frac{1}{2} \sqrt{x^{2} + y^{2}}$
Similarly, all other side are equal
$∴ P Q R S$ will be a rhombus.

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Similar questions

Q. P,Q,R,R, and S are respectively the mid-points of sides AB, BC, CD, And DA of a quadrilateral ABCD in which AC =BD, then PQRS is a rhombus.

Q. Question 5
P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and

A C ⊥ B D

. Prove that PQRS is a square.

A B C D

is a rectangular and

P, Q, R

and

S

are mid points of the sides

A B, B C, C D

and

D A

respectively. Show that the quadrilateral

P Q R S

is a rhombus .

Question 4
P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that $A C ⊥ B D$ . Prove that PQRS is a rectangle.

P, Q, R

and

S

are respectively the mid-points of the sides

A B, B C, C D

and

D A

of a quadrilateral

A B C D

such that

A C ⊥ B D

. Prove that

P Q R S

is a rectangle.

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P,Q,R and S are respectively the mid-points of the sides AB ,BC, CD and DA of a quarilateral ABCD in which AC= BD, prove that PQRS is a rhombus.

if AC=BDthen ABCD must be a square or a rectangleif ABCD is a rectangle,PB=x2, QB=y2∴ PQ=√PB2+QB2 (Pythagorus theorem)=√x24+y24PQ=12√x2+y2Similarly,AP=x2, AS=y2PS=√AP2+AS2=√x24+y24=12√x2+y2Similarly, all other side are equal∴ PQRS will be a rhombus.