Question

prove that the straight line joining the mid points of the sides of an isosceles trapezium in order form a rhombus

Answer 1

Let ABCD be the 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 is 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.