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Mathematics, 22.06.2021 22:20 depinedainstcom

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Mathematics, 20.06.2019 18:04
Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with āˆ dab = x, and āˆ abc = 90ā—¦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus āˆ pad = āˆ pbc. but pab is isosceles, hence āˆ pab = āˆ pba. subtracting, gives x = āˆ padāˆ’āˆ pab = āˆ pbc āˆ’āˆ pba = 90ā—¦. this is a preposterous conclusion ā€“ just where is the mistake in the "proof" and why does the argument break down there?
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