Absolutely beautiful proof of Alternate Segment Theorem. 🙂 When you created triangle EDB one just needs to look at where vertices D point of contact with Tangent Line FC. So angle Alpha DEB is readily available to us now which allows us too the magic of Triangle Sum Theorem Central Angle Theorem Angle Angle Similarity Theorem (For Ratios or Proportions) ...too determine X! Soooooooo Awesome! 🙂
In my opinion, the most fascinating aspect to this problem is that we can find the value of x without finding the lengths of any of the other line segments, nor the radius of the circle! In fact, it is possible to move point F farther away from the center of the circle, reconstruct the diagram, and, except for 18, 8 and x, all lengths will change, as well as the radius of the circle.
This was quite the challenge for one having long forgotten the alternate segment theorem. I finally decided I needed to find similar triangles and managed eventually to solve the puzzle and prove the theorem along the way (though not as elegantly).
Similarity of triangles: PB / DB = AB / EB x / DB = 18 / EB x /18 = DB / EB Similarity of triangles: PB / EB = BC / DB x / EB = 8 / DB 8 / x = DB / EB Equalling: x / 18 = 8 / x x² = 8 . 18 = 144 x = 12 cm ( Solved √ )
Assuming there is only one solution and the measure of the angle ∠ABC doesn't matter: Make ABC collinear using B as a hinge so that the diameter of the yellow circle is 18 + 8. Then apply the intersecting chords theorem: PB² = AB ⋅ BC ⇒ x² = 18 ⋅ 8 ⇒ x = √144 = 12.
OK I tried a bit of a strange approach using "similar quadrilaterals" (is that even a thing?) with ABPE and PBCD. The angles for each of them seem to match up the same as I arranged the lettering. I then used ratios and said 18/x = x/8. That gives me x = 12
Yes, I researched it and there are similar quadrilaterals. "Two quadrilaterals are similar quadrilaterals when the three corresponding angles are the same (the fourth angles automatically become the same as the interior angle sum is 360 degrees), and two adjacent sides have equal ratios." The "two adjacent sides" covers the special case of rectangles, which always meet the angle requirement but the base and height may not be in the same ratio, even if base and height are swapped on one rectangle. I don't know if there are any other special cases where the angle requirement is met but "adjacent sides" is not. Clearly, ABPE and PBCD have 2 right angles each, but can't have 4.
Hello. Actually I don't believe (I could be wrong) that there is a set value for alpha or the other angles or for the radius or for the length of ED or for the radius. It is fascinating that even with those things being variable, the distance from B perpendicular to the chord between the tangents will always be 12.
Absolutely beautiful proof of Alternate Segment Theorem. 🙂
When you created triangle EDB one just needs to look at where vertices D point of contact with Tangent Line FC.
So angle Alpha DEB is readily available to us now which allows us too the magic of Triangle Sum Theorem Central Angle Theorem
Angle Angle Similarity Theorem
(For Ratios or Proportions)
...too determine X!
Soooooooo Awesome! 🙂
Excellent!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
In my opinion, the most fascinating aspect to this problem is that we can find the value of x without finding the lengths of any of the other line segments, nor the radius of the circle! In fact, it is possible to move point F farther away from the center of the circle, reconstruct the diagram, and, except for 18, 8 and x, all lengths will change, as well as the radius of the circle.
Wonderful to see proof of Alternate Segment Theorem! Great teaching!
This was quite the challenge for one having long forgotten the alternate segment theorem. I finally decided I needed to find similar triangles and managed eventually to solve the puzzle and prove the theorem along the way (though not as elegantly).
Thank you for teaching
Similarity of triangles:
PB / DB = AB / EB
x / DB = 18 / EB
x /18 = DB / EB
Similarity of triangles:
PB / EB = BC / DB
x / EB = 8 / DB
8 / x = DB / EB
Equalling:
x / 18 = 8 / x
x² = 8 . 18 = 144
x = 12 cm ( Solved √ )
Thanks for video.Good luck sir!!!!!!!!!!!!
Thanks sir
Assuming there is only one solution and the measure of the angle ∠ABC doesn't matter:
Make ABC collinear using B as a hinge so that the diameter of the yellow circle is 18 + 8.
Then apply the intersecting chords theorem: PB² = AB ⋅ BC ⇒ x² = 18 ⋅ 8 ⇒ x = √144 = 12.
Simply great, Sir, many thanks!
Very nice solution
Amazing 👍
Thanks for sharing😊
Thank you! Cheers!
x=18sinα/sinβ=8sinβ/sinα ∴sinα/sinβ=2/3 ∴x=18(2/3)=12
Nice. With beta=angles BEA and BDP
Very neat...
Thank you! Cheers!
OK I tried a bit of a strange approach using "similar quadrilaterals" (is that even a thing?) with ABPE and PBCD. The angles for each of them seem to match up the same as I arranged the lettering. I then used ratios and said 18/x = x/8. That gives me x = 12
Yes, I researched it and there are similar quadrilaterals. "Two quadrilaterals are similar quadrilaterals when the three corresponding angles are the same (the fourth angles automatically become the same as the interior angle sum is 360 degrees), and two adjacent sides have equal ratios." The "two adjacent sides" covers the special case of rectangles, which always meet the angle requirement but the base and height may not be in the same ratio, even if base and height are swapped on one rectangle. I don't know if there are any other special cases where the angle requirement is met but "adjacent sides" is not. Clearly, ABPE and PBCD have 2 right angles each, but can't have 4.
@@jimlocke9320 So I lucked out then. Thanks.
What is the value of angle.Alfa,I fail to draw figure to scale
Hello. Actually I don't believe (I could be wrong) that there is a set value for alpha or the other angles or for the radius or for the length of ED or for the radius. It is fascinating that even with those things being variable, the distance from B perpendicular to the chord between the tangents will always be 12.
I am still in the box😮
congruent quadrilaterals
I am afraid 😢while listening thinking out of box.
No worries. We are all lifelong learners. That's what makes our life exciting and meaningful!
Cheers
Thanks sir...
If x is 12 then FD=?
asnwer=10cm isit hmm gmmm
Can I use similar quadrilateral AEPD,PDGB to find X?