An alternate simpler approach will be: after getting equation 1, consider triangle CFB, and get the equation 2 as, alpha + 2x = 90 deg.. then solve these two simultaneous equations.🙏🏼
To prove P of angle bisector CP is on BA not on BA extend: When P on is BA extend, congruent triangles CBP and CDP (SAS by construction) is false as their corresponding angles B and D are not equal, i.e. 2x not equal to 4x. This makes your method perfect.
Hello, I drew auxiliary line DE so that DE equals AD (and DC and CB) with E on line AB. THen you can show that angle CEB is 2x, therefore triangle CEB is isosceles and so EC=CB (and DE and DC). Thus triangle DEC is equilateral. You can then show that 5x=120.
I was able to solve it with sine law and trig lots of manipulations, but creating a triangle around the 4x angle and bisecting it to get 2 congruent triangles with each being congruent to a triangle that can be formed around the non-bisected 2x angle was truly a genius idea. (Even if you didn't fully imagine what can happen when you go after bisecting the 4x angle, the part about closing 1 isosceles triangle was obvious, just not obvious which one it should be, you bisect the 4x angle which got you a best case scenario of 2 right triangles, from there the next step of closing a right triangle around the 2x angle was obvious, but ending up with a 30-60-90 trianglewas such a beautiful coincidence, truely a beautiful solution for a beautiful puzzle)
Sir after getting that AC = 2CF .then Sin 30 = CF / ACthat is 1/2 we can obtain the equation 30 + alpha = 3x we can get the 2 nd easiest equation from any of the triangle AED,DEC,Or CFB as every triangle is rectangular triangle hence alpha + 2x will always be 90 . I think it is more easier to calculate the value of X,with the help of these two equation by replacing alpha The answer will be 5x=120 i.e x = 24
Per il teorema dei seni risulta √(2a^2-2a^2cos4x)/sin2x=a/sin(5x-90)...a lato uguale che si semplifica,rimane l'equazione in incognita x..risulta cos5x=(-1/2).x=24
Let's find x: . .. ... .... ..... First of all we add point E on AB such that ADE is an isosceles triangle (AD=DE). Then we can conclude: ∠AED = ∠DAE = 3x ⇒ ∠ADE = 180° − ∠DAE − ∠AED = 180° − 3x − 3x = 180° − 6x ⇒ ∠CDE = ∠ADC − ∠ADE = 4x − (180° − 6x) = 4x − 180° + 6x = 10x − 180° Since AD=DE and AD=CD, the triangle CDE is also an isosceles triangle (CD=DE), so we obtain: ∠DCE = ∠CED = (180° − ∠CDE)/2 = [180° − (10x − 180°)]/2 = (180° − 10x + 180°)/2 = (360° − 10x)/2 = 180° − 5x ⇒ ∠BEC = 180° − ∠AED − ∠CED = 180° − 3x − (180° − 5x) = 180° − 3x − 180° + 5x = 2x = ∠CBE The triangle BCE is an isosceles triangle as well and we finally obtain: CE = BC = CD = AD = DE ⇒ CD = CE = DE Therefore the triangle CED is not only an isosceles triangle, it is an equilateral triangle and we can conclude: ∠DCE = ∠CED = 180° − 5x = 60° ⇒ 120° = 5x ⇒ x = 24° ∠CDE = 10x − 180° = 60° ⇒ 240° = 10x ⇒ x = 24° ✓ Best regards from Germany
Very good.
Glad to hear that!
Thanks for the feedback ❤️
An alternate simpler approach will be: after getting equation 1, consider triangle CFB, and get the equation 2 as,
alpha + 2x = 90 deg.. then solve these two simultaneous equations.🙏🏼
Put point P on line AB so that
CP bisects the angle C.
Then there triangles ADP DCP
CBP are congruent.
Therefore angle C is 6x and
x is 24
To prove P of angle bisector CP is on BA not on BA extend: When P on is BA extend, congruent triangles CBP and CDP (SAS by construction) is false as their corresponding angles B and D are not equal, i.e. 2x not equal to 4x. This makes your method perfect.
sin(5x-90)=a/2a=1/2
5x-90=30
x=120/5=24
Very good
Thank you!
Hello, I drew auxiliary line DE so that DE equals AD (and DC and CB) with E on line AB. THen you can show that angle CEB is 2x, therefore triangle CEB is isosceles and so EC=CB (and DE and DC). Thus triangle DEC is equilateral. You can then show that 5x=120.
Great solution!
Ótima solução! 🎉🎉🎉
sin(3x)-sin(7x)-sin(2x)=0
Not sure of any good way to solve that, but it turns out to be correct.
Another way A=90-2x, A=3x-30 then 90-2x=3x-30, x=24
Difficult for me, I had to watch the video.
The proceeding shown is great 👍
Best wishes!
Good
I was able to solve it with sine law and trig lots of manipulations, but creating a triangle around the 4x angle and bisecting it to get 2 congruent triangles with each being congruent to a triangle that can be formed around the non-bisected 2x angle was truly a genius idea.
(Even if you didn't fully imagine what can happen when you go after bisecting the 4x angle, the part about closing 1 isosceles triangle was obvious, just not obvious which one it should be, you bisect the 4x angle which got you a best case scenario of 2 right triangles, from there the next step of closing a right triangle around the 2x angle was obvious, but ending up with a 30-60-90 trianglewas such a beautiful coincidence, truely a beautiful solution for a beautiful puzzle)
@ 2:35 , droppin that perpendicular auxiliary is genius! ...imoh 🙂.
Sir after getting that AC = 2CF .then Sin 30 = CF / ACthat is 1/2 we can obtain the equation 30 + alpha = 3x
we can get the 2 nd easiest equation from any of the triangle AED,DEC,Or CFB as every triangle is rectangular triangle hence alpha + 2x will always be 90 .
I think it is more easier to calculate the value of X,with the help of these two equation by replacing alpha The answer will be 5x=120 i.e x = 24
No hablo inglés. Pero explicas tan bien , que te entendí todo . 👍
👍👍👍
😮
It needs only elementary skills, but the way to find how combine them is not easy at all. I didn't find this time. Thank your for this nice puzzle !
I did not find as well.
Crude trig knowledge?
The aux lines were tricky but the algebra part became easy
360÷2+3+4+6_24
Per il teorema dei seni risulta √(2a^2-2a^2cos4x)/sin2x=a/sin(5x-90)...a lato uguale che si semplifica,rimane l'equazione in incognita x..risulta cos5x=(-1/2).x=24
شكرا لكم على المجهودات
0
Let's find x:
.
..
...
....
.....
First of all we add point E on AB such that ADE is an isosceles triangle (AD=DE). Then we can conclude:
∠AED = ∠DAE = 3x
⇒ ∠ADE = 180° − ∠DAE − ∠AED = 180° − 3x − 3x = 180° − 6x
⇒ ∠CDE = ∠ADC − ∠ADE = 4x − (180° − 6x) = 4x − 180° + 6x = 10x − 180°
Since AD=DE and AD=CD, the triangle CDE is also an isosceles triangle (CD=DE), so we obtain:
∠DCE = ∠CED = (180° − ∠CDE)/2 = [180° − (10x − 180°)]/2 = (180° − 10x + 180°)/2 = (360° − 10x)/2 = 180° − 5x
⇒ ∠BEC = 180° − ∠AED − ∠CED = 180° − 3x − (180° − 5x) = 180° − 3x − 180° + 5x = 2x = ∠CBE
The triangle BCE is an isosceles triangle as well and we finally obtain:
CE = BC = CD = AD = DE ⇒ CD = CE = DE
Therefore the triangle CED is not only an isosceles triangle, it is an equilateral triangle and we can conclude:
∠DCE = ∠CED = 180° − 5x = 60°
⇒ 120° = 5x
⇒ x = 24°
∠CDE = 10x − 180° = 60°
⇒ 240° = 10x
⇒ x = 24° ✓
Best regards from Germany
As always...terrific! 😇
@@dirklutz2818 Thanks a lot for your very kind feedback.
This mat is ronge