A polygon is called degenerate if one of its vertices falls on a line that joins its neighboring two vertices. In a pentagon ABCDE, AB = AE, BC = DE. P and Q are midpoints of AE and AB. PQ||CD, BD is perpendicular on both AB and DE. Prove that ABCDE is a degenerate pentagon.
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When you make a sketch (as good as you can) you observe that since BD is perp to AB and DE and AE = AB, that ABDE must be a square (not a rectangle). Now since CD is parallel to PQ and given that BC = DE = BD (square), triangle BCD will have to be isosceles. However, it can be verified that angles CBD and BDC are 45 degrees due to PQ with respect to AE and AB and that leads to a contradiction for C to be a distinct fifth point.
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