Show that the quadrilateral formed by joining the mid points of the pairs of consecutive sides of a rhombus is a rectangl
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Given AC,BD are diagonals of a quadrilateral ABCD are perpendicular.
P,Q,R and S are the mid points of AB,BC, CD and AD respectively.
Proof:
In ΔABC, P and Q are mid points of AB and BC respectively.
∴ PQ|| AC and PQ = ½AC ..................(1) (Mid point theorem)
Similarly in ΔACD, R and S are mid points of sides CD and AD respectively.
∴ SR||AC and SR = ½AC ...............(2) (Mid point theorem)
From (1) and (2), we get
PQ||SR and PQ = SR
Hence, PQRS is parallelogram ( pair of opposite sides is parallel and equal)
Now, RS || AC and QR || BD.
Also, AC ⊥ BD (Given)
∴RS ⊥ QR.
Thus, PQRS is a rectangle.
Hope this helped.:)
P,Q,R and S are the mid points of AB,BC, CD and AD respectively.
Proof:
In ΔABC, P and Q are mid points of AB and BC respectively.
∴ PQ|| AC and PQ = ½AC ..................(1) (Mid point theorem)
Similarly in ΔACD, R and S are mid points of sides CD and AD respectively.
∴ SR||AC and SR = ½AC ...............(2) (Mid point theorem)
From (1) and (2), we get
PQ||SR and PQ = SR
Hence, PQRS is parallelogram ( pair of opposite sides is parallel and equal)
Now, RS || AC and QR || BD.
Also, AC ⊥ BD (Given)
∴RS ⊥ QR.
Thus, PQRS is a rectangle.
Hope this helped.:)
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