By using the converse of bpt prove that the line joining the mid points of a non parallel sides of the trapezium are parallel to the parallel sides.

Dear Student,

Please find below the solution to the asked query:

We form our diagram , As :

Here , ABCD is a trapezium and AB  | | CD and by construction we form triangle ABE and M and N are mid points of non parallel sides AD and BC respectively .

In triangle ABE , AB  | | CD , So  by BPT we get

EDDA = ECCBEM - MDDA = EN - NCCBEMDA - MDDA = ENCB - NCCBEM2 MD - MD2 MD = EN2 NC - NC2 NC    (  M and N is mid point of AD and BC resepectively )EM2 MD - 12 = EN2 NC - 12EMMD = ENNCEMMA = ENNB
From Converse of BPT we get MN  | | AB , So

MN | | AB | | CD                                                           ( Hence proved )
Hope this information will clear your doubts about topic.

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