prove that the straight lines joining the midpoints of the sides of an isosceles trapezium in order form a of rhombus
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prove that the straight lines joining the midpoints of the sides of an isosceles trapezium in order form a of rhombus
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Answer:
Let ABCD is a trapezium.
AD=BC and AB is parallel to DC
Consider the triangles APS and PBQ.
AS=BQ
AP=PB
Angle SAB = angle ABQ
therefore, triangle SAP is congruent to triangle QBP
therefore, PS=PQ
we can similarly prove that triangle RDS to congruent to triangle RCQ.
Therefore, SR=RQ.
join PR
since the trapezium is an isosceles trapezium,
therefore PR will be a perpendicular bisector of
SQ. This can be proved by symmetry of the figure also.
Therefore, PQRS is a rhombus.
Step-by-step explanation:
sorry for late answer to your question.