Understand the problem
Source of the problem
Start with hints
We will now proceed towards proving in this direction.
Hence KHIM must be a parallelogram. Hence, KH || MI.
Also, \( \angle KHI =\angle HIG \). Hence, KH || EG.
Hence, it implies from KH || MI and KH || EG, that M,E,I,G are collinear.
Hence, MEF is isosceles.
Also, \( \angle MEF = \pi – \angle GEF = \pi – \angle EDA = \angle FDC \).
Hence, triangle MEF is similar to triangle FDC. This implies that \( \angle EFM = \angle DFC\).
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