Originally Posted by
Aquafina hi i have the following question:
this argument is given which makes an obtuse angle be a right angle, i have attached a picture of the shape.
Given the obtuse angle x, we make a quadrilateral ABCD with ANGLE DAB = x, and ANGLE ABC =90◦, andAD = BC. Say the perpendicular bisector toDC meets the perpendicular bisector to AB at P. Then PA = PB andPC = PD. So the triangles PAD and PBC have equal sides and are congruent. Thus ANGLE PAD = ANGLE PBC. But PAB is isosceles, hence ANGLE PAB = ANGLE PBA.
Subtracting, gives x = ANGLE PAD− ANGLE PAB = ANGLE PBC − ANGLE PBA = 90◦. This is a preposterous conclusion.
Where is the mistake in the “proof” and why does the argument break down there?