Problem 7
Let S and S’ be two circles that intersect at A and B. Let C be any point on S’. If CB intersects S at D, and the tangent at A to S’ intersects S at E, prove that:
AC || DE
Here’s a simulation that demonstrates the problem.
You can drag the circles S and S’ and observe AC and DE. You can also drag the point C. Is AC always parallel to DE?
Solution
To solve this problem, you’ll need to know three simple theorems.
Parallel Lines and Transversals
If a line (transveral) intersects two parallel lines, then the co-interior angles are supplementary. That is, they always add up to 180°. Here’s a simulation that demonstrates this property.
You can drag each point and observe the sum of the co-interior angles. Is it always 180°?
Conversely, if a transveral intersects two lines and the co-interior angles are supplementary, then the two lines must be parallel.
Cyclic Quadrilaterals
The opposite angles of a quadrilateral are supplementary. Here’s a simulation that demonstrates this property.
You can drag each vertex and observe the sum of the opposite angles. Is it always 180°?
Alternate Segment Theorem
If a tangent is drawn to a circle at the end point of its chord PQ, then the angle between the tangent and PQ is the same as PQ subtends in the alternate segment of the circle.
That was too much information packed into once sentence. Have a look at the simulation to understand this theorem.
You can drag the points P, Q, and R. Is the angle between PQ and the tangent same as the angle subtended in the alternate segment?
Now, to solve the problem, you need to look for co-interior angles and prove them as supplementary. To do that, you’ll need to apply the alternate segment theorem (look for the tangent), as well as the cyclic quadilateral property.
That’s all for now. I suggest you proceed on your own from this point. Good luck!
Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Facebook page.
Source: Math Club
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