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Problem(Plane Geometry)

TheFurret Wed 30 Sep 19:35:44 2020 No.12178920 View Reply Original Report

Quoted By: >>12179246 >>12179254 >>12179439 >>12179931 >>12179999 >>12180035

Hello there guys, I've been trying to solve this problem with triangles similarity, but I'm not quite sure of how to start it.

Anonymous

Anonymous Wed 30 Sep 2020 19:44:35 No.12178948 Report

Quoted By: >>12179025 >>12179252

Are you trying to solve for x?
Here's an intermediate goal.
Compute AC as a function of alpha and ED.
Compute x as a function of AC and alpha.
Hope that alpha drops out.

Anonymous

Anonymous Wed 30 Sep 2020 20:09:50 No.12179025 Report

Quoted By: >>12179029 >>12179226 >>12179585

>>12178948
That method would use trigonometry. A pure Euclidean-style construction might go as follows:
Draw a circle whose center is the midpoint of AC. It will go through B and D.
Extend the angle bisector of BAC past BC to meet the circle in a new point F.
You can show CF = CD = BF.
You can also show angle BCF = alpha.
Now connect F to the midpoint of BC, and call that point G.
FCG is a right triangle, and is congruent to CDE.
In particular, DE = CG.
Now x should be obvious.

Anonymous

Anonymous Wed 30 Sep 2020 20:11:32 No.12179029 Report

Quoted By:

>>12179025
*Note that angle EDC equals alpha.

Anonymous

Anonymous Wed 30 Sep 2020 21:12:17 No.12179226 Report

Quoted By: >>12179383 >>12179399 >>12179410 >>12179489

>>12179025
>Draw a circle whose center is the midpoint of AC. It will go through B and D.

Sure...

Anonymous

Anonymous Wed 30 Sep 2020 21:16:39 No.12179246 Report

Quoted By: >>12179252

>>12178920
A solution will exists for alphas <60deg. The answer will be a function of alpha. You can calculate AC using trigonometry and similarity with the triangle of known side. Then, you can use trigonometry and the other angle.

Anonymous

Anonymous Wed 30 Sep 2020 21:18:42 No.12179252 Report

Quoted By: >>12179371

>>12178948
>>12179246
Same answers, really. Silly remark in the first one, as alpha won't drop out. x will be much different with different alphas.

Anonymous

Anonymous Wed 30 Sep 2020 21:19:16 No.12179254 Report

Quoted By:

>>12178920
>ABC = ADC = 90
But they are clearly not 90 in your picture. This is a good starting point.

Anonymous

Anonymous Wed 30 Sep 2020 21:44:09 No.12179371 Report

Quoted By:

>>12179252
>x will be much different with different alphas
Try it.

Anonymous

Anonymous Wed 30 Sep 2020 21:46:33 No.12179383 Report

Quoted By:

>>12179226
For any right triangle, the circumscribed circle has its center at the midpoint of the hypotenuse.

Anonymous

Anonymous Wed 30 Sep 2020 21:50:25 No.12179399 Report

Quoted By:

>>12179226
That was supposed to also go through A and C. This works because you ABC and ADC are right triangles.

Anonymous

Anonymous Wed 30 Sep 2020 21:53:06 No.12179410 Report

Quoted By:

>>12179226
This was supposed to be a circle with that center going through A and C. Because ABC and ADC are right triangles, this circle goes through B and D.

Anonymous

Anonymous Wed 30 Sep 2020 21:59:54 No.12179439 Report

Quoted By:

>>12178920
I would start at AED = 90 degree.

Anonymous

Anonymous Wed 30 Sep 2020 22:15:00 No.12179489 Report

Quoted By: >>12181241

>>12179226
https://www.codecogs.com/library/maths/geometry/pure/triangles-and-circles.php
Theorem g.
Proof trivial and left as exercise

Anonymous

Anonymous Wed 30 Sep 2020 22:40:29 No.12179585 Report

Quoted By: >>12179627

>>12179025
>CF = CD = BF
CF = DF = BF != CD

Anonymous

Anonymous Wed 30 Sep 2020 22:49:44 No.12179627 Report

Quoted By: >>12179791

>>12179585
But A, B, F, C, and D are all on the circle, and angles BAF, FAC, and CAD are all equal, so BF, FC, and CD are all equal.

Anonymous

Anonymous Wed 30 Sep 2020 23:30:55 No.12179791 Report

Quoted By:

>>12179627
I'm special. Missed where F was meant to go.

Anonymous

Anonymous Thu 01 Oct 2020 00:31:51 No.12179931 Report

Quoted By:

>>12178920
This figure is wrong, BAD and BCD both need to be obtuse angles

Anonymous

Anonymous Thu 01 Oct 2020 00:56:27 No.12179999 Report

Quoted By:

>>12178920
This construction isn’t unique.

Anonymous

Anonymous Thu 01 Oct 2020 01:07:56 No.12180035 Report

Quoted By:

>>12178920
Draw another line, the BD, and everything unfolds with pythagorean theorem expanded of cosinus and not just right angle.

Anonymous

Anonymous Thu 01 Oct 2020 07:54:45 No.12181241 Report

Quoted By: >>12181567

>>12179489
A and C are not necessarily on a circle drawn to pass through B and D with a center at the midpoint between A and C, theorem g doesn't apply

Anonymous

Anonymous Thu 01 Oct 2020 11:22:53 No.12181567 Report

Quoted By: >>12181738

>>12181241
B & D are right angles, as per definition, so theorem g applies, and all 4 points, A, B, C & D are on the same circle.

Anonymous

Anonymous Thu 01 Oct 2020 12:33:31 No.12181738 Report

Quoted By: >>12182689

>>12181567
>and all 4 points, A, B, C & D are on the same circle.
Where are you getting this

Anonymous

Anonymous Thu 01 Oct 2020 17:49:56 No.12182689 Report

Quoted By:

>>12181738
Going with "theorem g" (it's got an actual greek name I don't remember right now), if B is 90 degrees, then there exists a circle that touches A, B & C, with a center between A & C.
If D is 90 degrees, there exists a circle that touches A, C & D, with centre between A & C.
So, there are two circles with exactly the same diameter (AC) and touch at two points (A & C). So they have to be the same circle, and that circle touches all 4 points, A, B, C & D.
Ignore picture in OP, it's garbage.

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