I remember doing this about a year ago and I hated it so much that I feared clicking on this video. But for some reason, it feels so easy. Eddie, you're a wonderful teacher.
I might have a good proof of opposite angles in a cyclic quadrilateral adding up to 180 degrees such that the center of the circle is completely outside that quadrilateral. Let's say we go counterclockwise from the top of the circle and put points A, B, C, and D of the quadrilateral on the circle so that we get a cyclic quadrilateral that is completely outside the center of the circle. Next, we rotate everything so that segment AD is horizonal and under the center of the circle. Then we draw an imaginary vertical line from the center of the circle to its base and call that segment OT. If the points B and C are both either to the left of or on the imaginary line OT then re-position point A to the top of the circle. Angles BAD and BCD will still be the same. Angle ABC will get smaller by a quantity f while angle ADC will get larger by the same quantity f. Since we already proved that a cyclic quadrilateral with the circle's center inside of it has both pairs of its opposite angles adding to 180 degrees then we can say that the original quadrilateral with the circle's center outside of it is also a cyclic quadrilateral since one pair of opposite angles is the same as one of the pairs of opposite angles of the converted quadrilateral that completely surrounds the circle's center and one of the angles in the other pair of opposite angles is f more than its corresponding angle while the other is f less than its corresponding angle. Since the f more and f less cancel each other out then the sum of those angles are equal to the sum of the corresponding angles of the converted quadrilateral. Thus, the quadrilateral whose vertices are all outside the circle's center must be a cyclic quadrilateral, too. If the points B and C were both either to the right of point T or at point T then you would move point D to the top of the circle. With D at the top of the circle, the roles of the two pairs of opposite angles would switch places, but the proof would be the same. If B is to the left of T while C is to the right of T then you can move either A or D to the top of the circle and prove accordingly.
Here in germany at least, the naming of an angle is highly dependent of the order. In the drawing, ∠AOC would be be reflex angle and ∠COA the obtuse one.
I remember doing this about a year ago and I hated it so much that I feared clicking on this video. But for some reason, it feels so easy. Eddie, you're a wonderful teacher.
I wish you were my maths teacher I hope you get the Nobel prize because you deserve it!
I might have a good proof of opposite angles in a cyclic quadrilateral adding up to 180 degrees such that the center of the circle is completely outside that quadrilateral. Let's say we go counterclockwise from the top of the circle and put points A, B, C, and D of the quadrilateral on the circle so that we get a cyclic quadrilateral that is completely outside the center of the circle. Next, we rotate everything so that segment AD is horizonal and under the center of the circle. Then we draw an imaginary vertical line from the center of the circle to its base and call that segment OT. If the points B and C are both either to the left of or on the imaginary line OT then re-position point A to the top of the circle. Angles BAD and BCD will still be the same. Angle ABC will get smaller by a quantity f while angle ADC will get larger by the same quantity f. Since we already proved that a cyclic quadrilateral with the circle's center inside of it has both pairs of its opposite angles adding to 180 degrees then we can say that the original quadrilateral with the circle's center outside of it is also a cyclic quadrilateral since one pair of opposite angles is the same as one of the pairs of opposite angles of the converted quadrilateral that completely surrounds the circle's center and one of the angles in the other pair of opposite angles is f more than its corresponding angle while the other is f less than its corresponding angle. Since the f more and f less cancel each other out then the sum of those angles are equal to the sum of the corresponding angles of the converted quadrilateral. Thus, the quadrilateral whose vertices are all outside the circle's center must be a cyclic quadrilateral, too. If the points B and C were both either to the right of point T or at point T then you would move point D to the top of the circle. With D at the top of the circle, the roles of the two pairs of opposite angles would switch places, but the proof would be the same. If B is to the left of T while C is to the right of T then you can move either A or D to the top of the circle and prove accordingly.
Here in germany at least, the naming of an angle is highly dependent of the order. In the drawing, ∠AOC would be be reflex angle and ∠COA the obtuse one.
Will u make videos on advanced maths topics like
Vector calculus
LDE
Partial derivatives
Etc
I have studied as
The exterior angle of a triangle is the sum of REMOTE INTERIOR ANGLES
Why u didn't use the word REMOTE?
it's 1:00 am here in India, and today is my maths exam. I'll be studying whole night!!
vedant haha, same here! But My math test was yesterday😃