Hi guys ,
Two unequal circles centres A and B intersect at X, Y, prove that AB bisects XY at right angles.
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Answer:
When you are taught to construct the perpendicular bisector of a segment PQ, you draw arcs with radius R, centered at P and Q so that they intersect at point A. Then you draw arcs with radius R, centered at P and Q so that they intersect at point C, on the opposite side of segment PQ from point A. You could instead draw draw arcs with radius r, centered at P and Q so that they intersect at point B, on the opposite side of segment PQ from point A. In the first case, APQC is a rhombus. In the second case, if r is different from R, APQB is a kite. I do not memorize textbook proofs that seem obvious to me, so if I had to give a proof, it would be the explanation below. If I was taking a geometry course and I needed a “two-column proof” I would look up the jargon used in textbook or class notes so I could write the expected phrases, such as “AM is congruent to AM by …”
If you consider point M, the midpoint of PQ, you realize that triangles APM and AQM are congruent, because they have congruent pairs of sides. Side AM is a side of both triangles, and is obviouly congruent with inself, so it counts as a pair of congruent sides. PM and QM are congruent because M is the midpoint, and by the definition of midpoint. Sides AP and AQ are congruent because they are radii of the same circle. As triangles APM and AQM are congruent, their corresponding parts are congruent, and in math classes, they abbreviate that theorem as CPCTC for (Corresponding Parts of Congruent Triangles are Congruent). In particular, corresponding angles PMA and QMA are congruent. As they both add to a flat (180 degree) angle, their measure must be half of that. They are right angles, so AM is part of the line that is the perpendicular bisector of PQ. The same can reasoning can be used to prove that BM is part of the line that is the perpendicular bisector of PQ. AM and BM are both part of the same line, the only line that is the perpendicular bisector of PQ. That line can be called AB, and that is what you would call it if you had not met point M. So Ab mot only bisects PQ, it is also perpendicular to PQ.
Answer:
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