Question

prove that the straight lines joining the midpoints of the sides of an isosceles trapezium in order form a of rhombus​

Answer 1

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.

Answer 2

Step-by-step explanation:

prove that the straight lines joining mid points of sides of an isosceles trapezium taken in order form a rhombus. Angle SAB = Angle ABQ. 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.