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Solution :

`AL= LB & CM = MD ` <br>
`AB _|_ OL and CD _|_ OM` <br>
In `/_\ LOX and /_\ MOX` <br>
`AB= CD & OL= OM` <br>
`/_ OLX = /_ OMX = 90^O` <br>
`OX= OX `common side in both triangles <br>
`/_\ LOX ~= /_\ MOX` <br>
`LX= MX` eqn(1) <br>
`AB=CD` eqn (3)<br>
`AB/2 = CD/2`<br>
`So, BL= CM` eqn(2)<br>
add eqn 1 & 2 <br>
`LX + BL = MX + CM` <br>
`BX = CX` eqn (4)<br>
subtract eqn 4 from 3
`AB - BX = CD- CX` <br>
`AX= DX`hence proved