In the figure, Segment AB is a diameter of the smaller of the two concentric circles. Segments AP and BQ are tangent to the smaller circle at A and B, respectively.
PROVE that Segments AB and PQ intersect at the center of the circle.
For clarity, I made the circles very non-concentric in the following picture.
Let O be the center of the circles, by definition. Then AP = BQ by the Pythagoras theorem. Therefore, $\displaystyle \triangle AOP= \triangle BOQ$. So, $\displaystyle \angle AOP=\angle BOQ$, which means that POQ is a straight line.
I did this proof as follows
given two concentric circles,AB diameter of small circle, AP and BQ tangents to small circle @A and B
1 extend PA and BQ to C and D points on the large circle
2 TAngents @ A and B meet diameter AB @ 90 deg property of tangents
3 PAC and QBD are parallel two lines perpendicular to a third line are parallel
4 PAC and QBD are equal in lenght Each is a chord of large circle perpendicular to AB and equidistant to center of circles m the midpoint of AB
5 PCQD is a rectangle CQ=PD =AB perpendicular distance between parallel are equal
6 QP and CD are diagonals of the rectangle meeting AB @ m the middle of rectangle Property of rect diagonals
7 QM =MP CM=DM AM- MB Mis the midpoint of AB and the point bisecting the three lines QP, CD, AB
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