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

$$\begin{gathered} \frac{\mathrm{ED}}{\mathrm{DA}} = \frac{\mathrm{EC}}{\mathrm{CB}} \\ \Rightarrow \frac{\mathrm{EM} - \mathrm{MD}}{\mathrm{DA}} = \frac{\mathrm{EN} - \mathrm{NC}}{\mathrm{CB}} \\ \Rightarrow \frac{\mathrm{EM}}{\mathrm{DA}} - \frac{\mathrm{MD}}{\mathrm{DA}} = \frac{\mathrm{EN}}{\mathrm{CB}} - \frac{\mathrm{NC}}{\mathrm{CB}} \\ \Rightarrow \frac{\mathrm{EM}}{2 \mathrm{MD}} - \frac{\mathrm{MD}}{2 \mathrm{MD}} = \frac{\mathrm{EN}}{2 \mathrm{NC}} - \frac{\mathrm{NC}}{2 \mathrm{NC}} ( \mathrm{M} \mathrm{and} \mathrm{N} \mathrm{is} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AD} \mathrm{and} \mathrm{BC} \mathrm{resepectively} ) \\ \Rightarrow \frac{\mathrm{EM}}{2 \mathrm{MD}} - \frac{1}{2} = \frac{\mathrm{EN}}{2 \mathrm{NC}} - \frac{1}{2} \\ \Rightarrow \frac{\mathrm{EM}}{\mathrm{MD}} = \frac{\mathrm{EN}}{\mathrm{NC}} \\ \mathbf{\Rightarrow }\frac{\mathbf{EM}}{\mathbf{MA}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{EN}}{\mathbf{NB}} \end{gathered}$$
From Converse of BPT we get MN  | | AB , So

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

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards