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• Module 2 Day 8 Challenge Part 1

  Arc Measure and Congruent Inscribed Angles

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  @victorioussheep Good questions! The "two angles that are missing" are these two, marked in yellow, which do turn out to be 60 degrees (I think Prof. Loh also points to them in the video, but I see how it could be a bit unclear).
  missingangles.png

  Solving for those yellow angles, rather than the red ones you marked, is a good way to use what you know about inscribed angles and cyclic quadrilaterals. Specifically, Prof. Loh uses the fact that an inscribed angle's measure is half the measure of the arc it cuts off.

  But you could definitely solve for the red angles first! Take the triangle with angles 20, 60, and (unknown red angle). The unknown red angle is equal to 100 degrees because the angles of a triangle add to 180. Then, the supplementary red angle is 80 degrees (because 100+80 = 180). Finally, the last red angle equals 180-40-80 = 60 degrees (using the triangle with angle 40, 80, (unknown red/blue angle)), like we found before.

  So basically, there are multiple valid ways to solve the problem, and each way emphasizes different methods 🙂 finding the yellow angles first is most useful for illustrating the usefulness of inscribed angles, but you could solve it either way.

• Module 2 Day 8 Challenge Part 2

  Opposite Angles of a Cyclic Quadrilateral

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• Module 2 Day 8 Challenge Part 3

  Supplementary Angles, Counting by Casework

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  @debbie
  Prof Loh does it in Module 3, Day 4 Challenge Question 🙂

• Module 2 Day 8 Challenge Part 4

  Proving That a Quadrilateral is Cyclic

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  N

  Hi @The-Rogue-Blade!

  We assumed that there exist four points that have equal angles as shown on picture below, and we want to prove that these four points lie on the one circle by contradiction.

  M2D8C4-2.jpg

  We know that inscribed angles that subtend the same arc on the circle are equal. So we are drawing an additional line and get one more green angle that equal previous two:

  M2D8C4.jpg

  So now we have a triangle with interior and exterior angles that are equal. Hence, those two points are the same ones. 🙂

• Module 2 Day 8 Your Turn Part 1

  Point to Circle With Secant Lines, Similar Triangles

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• Module 2 Day 8 Your Turn Part 2

  Power of a Point For Point Outside Circle

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  @The-Rogue-Blade This is a good question! If we take the Power of a Point Theorem literally, then the segments that we are multiplying together should live up to the name "Power of a Point"; they should all emanate from the same point.

  M2W2D8-y-part-2-power-of-point-why-not-ae-times-af-equals-bc-squared.png

  In the above illustration, the correct answer is shown:

  $$ CF \times CE = (BC)^2 $$

  Now consider \(\overline{AE}\) and \(\overline{AF};\) these segments emanate from point \(A,\) but the segment \(\overline{BC}\) emanates from a \(\textcolor{red}{\text{different}}\) point, \(C.\) We have not one point, but \(2\) points.

  M2W2D8-y-part-2-power-of-point-why-not-ae-times-af-equals-bc-squared2.png

  Unfortunately, it's not called "Power of Two Points," so

  $$ AE \times AF \neq (BC)^2 $$

  🙂