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Given : `square ABCD` is a rectangle in which P,Q, R and S are the mid-point of AB, BC, CD and DA respectively. <br> To prove : `square PQRS` is a rhombus. <br> Construction : Join AC. <br> Proof. First of all we will prove,`square PQRS` is a parallelogram . Then two consecutive sides of a parallelogram are equal i.e., PS=PQ. In this way we can get a rhombus. <br> In `triangle`, since P and Q ar the mid-points of AB and BC respectively. <br> `therefore PQ"||"AC and PQ=(1)/(2)AC" "` ("mid-point theorem")...(1) <br> Now ,in `triangleADC,` sicne S and R are the mid-points of AD and DC respectively. <br> `therefore SR"||"AC and SR=(1)/(2)AC" "(mid-point theorem)...(2)` <br> `therefore` From (1) adn (2) , we get <br> `PQ"||"SR and PQ=SR` <br> Thus, in equilateral PQRS, one pair of opposite sides in parallel sides in parallel and equal . <br> `therefore square PQRS` is a parallelogram.`" " ...(3)` <br> Since, `AD=BC" " `(opposite sides of a rectangle) <br> `implies (1)/(2)AD =(1)/(2)BC` <br> `implies SA=QB" "`(`because` S and Q are the mid-points) <br> Thus, in `triangleSAP, and triangleQBP` <br> `therefore {{:(SA=QB" "("just proved")),(angle1=angle2" "("each " 90^(@) "as angle of rectangle")),(AP=BP" "(therefore "P is the mid -point of AB")):}` <br> `therefore " "triangleSAP~=triangleQBP, " "` (SAS criterion of congruence) <br> `implies PS=PQ" "`(c.p.c.t)...(4) <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_IX_C08_S01_025_S01.png" width="80%">