Please help me with this prove.
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Are AB and CD equal?
Let $\displaystyle AC\cap BD=O$ and $\displaystyle MN\perp BC, \ M\in AD, \ N\in BC, \ O\in MN$ Then $\displaystyle DH=MN$ and $\displaystyle MO=\frac{AD}{2}, \ ON=\frac{BC}{2}$ $\displaystyle \Rightarrow MN=MO+ON=\frac{AD+BC}{2}=EF$
Thank you for your answer. I don't understand this. Please give me a simple prove.
See attachment.
Last edited by malaygoel; Jun 27th 2009 at 12:05 AM.
Thanks a lot! I also spent sometime thinking about this problem again and found an another way of solving. Triangle BDG is a right isosceles triangle. ADGC is a parallelogram.
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