0:21 Draw a line from C to point E on AB such that CE bisected angle DCB. Triangles CDE and CEB are congruent. Therefore angle CDE = 9X. Triangles ADE and EDC are congruent. Therefore angle ADE= 9X. As sum of all the angles of the quadrilateral = 360 45X=360 and therefore X=8
I constructed the same CE (angle bisector) but then noted CDE is congruent with ADE, as well as CBE. Then the three angles at E are congruent and add to 180. so angle CEB (and corresponding in each triangle) is 60, leaving 9x+6x =15x = 120 as the remaining angle sum in each triangle.
A much easier approach is to draw from C to the base AB making 9X angle. Then connect the line to D , You will end up with an equilateral triangle each angle is 60 degrees which is equal to 180- 15 X. Nice problem
It is unimaginable how can you design such amazing puzzle. I can solve it by sitting down with a pen and long computation. Joining CE where E at AB with angle CEB=9x, and then joining DE too, so we can show that angle DEA is 6x too, and thus triangle CDE is equilateral triangle, ...finally we can get x=8.
We know that the four angles must equal 360. 12x+9x+6x=27x Assuming the diagram, as usual is not necessary to scale and only the marked dimensions are "current", then if x=10 than the missing angle is 90, or 9x Other possibilities, if x=9 than the missing value is 15x. So, x is between 9 and 10, probably 9 based on the image, and thus the angle is 135°
What's the story with all those related angles? They would not exist in reality, 6x vs 12x, etc. So what is the theoretical solution for the problem? Please give an answer!
Lavorando col teorema dei seni risulta cos13, 5xsin9x=sin27xcos7,5x..e con un po' di pazienza risulta una soluzione per x=30.. che però non è accettabile, mi pare
I think I cheated, but I ended up with the correct answer. Angle ADC = 360-27x, so x MUST be less than 14 (360 - 27 * 14 is a negative number). I did a table analysis starting with x = 5, and then increased by one unit until all the values equaled 360, which is the sum of all of the interior angles. 5, 6 and 7 did not work. x = 8 did work, so x = 8.
You can’t claim ASA with what you have presented. You need to prove angle ADF is equivalent to angle CDF. That’s easily done using the sum of angles in a triangle theorem. But you skipped over an important part of the proof. Geometry students would lose points on their proof!
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Your approach so nice.
0:21 Draw a line from C to point E on AB such that CE bisected angle DCB. Triangles CDE and CEB are congruent. Therefore angle CDE = 9X. Triangles ADE and EDC are congruent. Therefore angle ADE= 9X. As sum of all the angles of the quadrilateral = 360
45X=360 and therefore X=8
I constructed the same CE (angle bisector) but then noted CDE is congruent with ADE, as well as CBE. Then the three angles at E are congruent and add to 180. so angle CEB (and corresponding in each triangle) is 60, leaving 9x+6x =15x = 120 as the remaining angle sum in each triangle.
I did the same - it works out very neatly and quick.
A much easier approach is to draw from C to the base AB making 9X angle. Then connect the line to D , You will end up with an equilateral triangle
each angle is 60 degrees which is equal to 180- 15 X. Nice problem
Excellent! Your problems are great and your solutions are beautiful. This is the best place for synthetic geometry on youtube.
Thank you! Cheers!
It is unimaginable how can you design such amazing puzzle. I can solve it by sitting down with a pen and long computation. Joining CE where E at AB with angle CEB=9x, and then joining DE too, so we can show that angle DEA is 6x too, and thus triangle CDE is equilateral triangle, ...finally we can get x=8.
Thank you for your sharing Proffesor
Thanks for video.Good luck sir!!!!!!!!!!
Thank you too
Complicated diagram
Marvelous solution
Thanks a lot
Thank you dear! Cheers! 😀
At about 6:19, line segment CE disappears. However, we can get the solution faster if we don't delete CE and consider ΔCEB. We note that
Your professional passion for math is my pleasure. 🙂
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Draw a parallel to side AB through point C and notice that 12x + 9x < 180, ie less than 8.57. This might help in a multiple choice test.
thank you for wonderful math solution
You are most welcome
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In triangle DEC, sin(6x) = DE/DC.
In triangle ABD,
1. angle ABD= 15x - 90
2. sin (6x)/ sin(15x - 90) = DB/AD = 2sin(6x).
Thus 15x-90=30.
Great solution.
Thank you! Cheers! 😀
3:02 ASA means the side is touched by both angles, not across from it! the side is included
You said it was ASA but isn't it actually AAS? Because the side was not bracketed by the two angles.
I agree. AAS
Thank you, Sir, for another wonderful video!❤🥂
So nice of you, my dear friend😀
We know that the four angles must equal 360.
12x+9x+6x=27x
Assuming the diagram, as usual is not necessary to scale and only the marked dimensions are "current", then if x=10 than the missing angle is 90, or 9x
Other possibilities, if x=9 than the missing value is 15x.
So, x is between 9 and 10, probably 9 based on the image, and thus the angle is 135°
After seeing the Angle Addition Postulate for obtaining 9x-30°, the problem became easy.
Thank you! Cheers! 😀
What's the story with all those related angles? They would not exist in reality, 6x vs 12x, etc. So what is the theoretical solution for the problem? Please give an answer!
The total of the angles of the quadrilatère is 360°
#6:15 , At the "6 munite 15 second" triangle : ( 6x ) + ( 9x-30 ) + 90 = 180 => x = 8
thanks you so much sir
Most welcome
Lavorando col teorema dei seni risulta cos13, 5xsin9x=sin27xcos7,5x..e con un po' di pazienza risulta una soluzione per x=30.. che però non è accettabile, mi pare
Divide
I think I cheated, but I ended up with the correct answer. Angle ADC = 360-27x, so x MUST be less than 14 (360 - 27 * 14 is a negative number). I did a table analysis starting with x = 5, and then increased by one unit until all the values equaled 360, which is the sum of all of the interior angles. 5, 6 and 7 did not work. x = 8 did work, so x = 8.
Thank you! Cheers! 😀
The puzzle comes a bit earlier😉. Let y be the angle D, so y+27x=360, then .....🙄🤨😅
Thank you! Cheers! 😀
Good night sir
Same to you dear
By the way, ∠CAB = ∠DBA. Simply because ∠ADC is twice ∠ABC just like ∠BCD is twice ∠BAD.
With the median CE being perpendicular, you had enough info already to calculate x
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X = 8
X is 8
You can’t claim ASA with what you have presented. You need to prove angle ADF is equivalent to angle CDF. That’s easily done using the sum of angles in a triangle theorem. But you skipped over an important part of the proof. Geometry students would lose points on their proof!