pqrs is trapezium in which pqllrs and ps=qr lf A,B,C,D are mid point of pq,qs,rs,and pr respectively then show that ABCD is a rhombus
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the base angles A and B are equal.
The midpoints of AB, BC,CD and DA are P,Q,R and S respectively.
By the definition of isosceles trapezium, the common perpendicular bisector to the bases of isosceles trapezoid divides the trapezium in to two congruent trapezoids.
Here PR is the line joining the midpoints of the bases and hence it is the perpendicular bisector of the bases.
Therefore, the diagonal PR of the quadrilateral PQRS bisects AB and CD perpendicularly.
In the same way, QS is the line joining the midpoints of the sides and hence it is the bisector.
Since AB is parallel to CD and QS is the line joining the midpoints Q and S, we have QS is parallel to AB.
Since PR is the perpendicular biscetor to the sides AB and CD, PR bisects QS, the line parallel to AB.
Since the diagonals of the quadrilateral bisects perpendicularly, the quadrilateral is a rhombus.
Thus, PQRS is a rhombus.
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The midpoints of AB, BC,CD and DA are P,Q,R and S respectively.
By the definition of isosceles trapezium, the common perpendicular bisector to the bases of isosceles trapezoid divides the trapezium in to two congruent trapezoids.
Here PR is the line joining the midpoints of the bases and hence it is the perpendicular bisector of the bases.
Therefore, the diagonal PR of the quadrilateral PQRS bisects AB and CD perpendicularly.
In the same way, QS is the line joining the midpoints of the sides and hence it is the bisector.
Since AB is parallel to CD and QS is the line joining the midpoints Q and S, we have QS is parallel to AB.
Since PR is the perpendicular biscetor to the sides AB and CD, PR bisects QS, the line parallel to AB.
Since the diagonals of the quadrilateral bisects perpendicularly, the quadrilateral is a rhombus.
Thus, PQRS is a rhombus.
plsss mark it as the brainliest
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