Mixing
Geometry with triangle.
$begingroup$
Any other solutions(advice) are welcome.
$angle BAC=60^circ, ;;;angle ACB=x,;;; overline {BD}=overline{BC}=overline{CE} $
geometry triangle angle
asked Jan 22 at 11:17
mina_worldmina_world
1799
$endgroup$
add a comment |
$begingroup$
Any other solutions(advice) are welcome.
$angle BAC=60^circ, ;;;angle ACB=x,;;; overline {BD}=overline{BC}=overline{CE} $
geometry triangle angle
asked Jan 22 at 11:17
mina_worldmina_world
1799
$endgroup$
1
$begingroup$
If you want to see a solution that is different from the image you provided, it is better to describe what direction or type of approach you'd like to see. Basically if you don't show your own thoughts and efforts, the post will most likely be ignored or closed by community votes.
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:29
add a comment |
$begingroup$
Any other solutions(advice) are welcome.
$angle BAC=60^circ, ;;;angle ACB=x,;;; overline {BD}=overline{BC}=overline{CE} $
geometry triangle angle
asked Jan 22 at 11:17
mina_worldmina_world
1799
$endgroup$
Any other solutions(advice) are welcome.
$angle BAC=60^circ, ;;;angle ACB=x,;;; overline {BD}=overline{BC}=overline{CE} $
geometry triangle angle
geometry triangle angle
asked Jan 22 at 11:17
mina_worldmina_world
1799
asked Jan 22 at 11:17
mina_worldmina_world
1799
edited Jan 22 at 17:06
mina_world
asked Jan 22 at 11:17
mina_worldmina_world
1799
asked Jan 22 at 11:17
mina_worldmina_world
1799
asked Jan 22 at 11:17
mina_worldmina_world
1799
1799
1
$begingroup$
If you want to see a solution that is different from the image you provided, it is better to describe what direction or type of approach you'd like to see. Basically if you don't show your own thoughts and efforts, the post will most likely be ignored or closed by community votes.
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:29
add a comment |
1
$begingroup$
If you want to see a solution that is different from the image you provided, it is better to describe what direction or type of approach you'd like to see. Basically if you don't show your own thoughts and efforts, the post will most likely be ignored or closed by community votes.
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:29
1
1
$begingroup$
If you want to see a solution that is different from the image you provided, it is better to describe what direction or type of approach you'd like to see. Basically if you don't show your own thoughts and efforts, the post will most likely be ignored or closed by community votes.
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:29
$begingroup$
If you want to see a solution that is different from the image you provided, it is better to describe what direction or type of approach you'd like to see. Basically if you don't show your own thoughts and efforts, the post will most likely be ignored or closed by community votes.
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$angle CBA=120^circ-x$$
Triangle $DBC$ is isosceles:
$$angle CDB=angle DCB=frac12angle CBA=60^circ-frac x2$$
Triangle $BCE$ is isosceles:
$$angle CEB=angle CBE=frac12angle BCA=frac x2$$
Now, from triangle:
$$angle BFC=120^circ$$
Take a look at quadrilateral $ABFC$: The sum of opposite angles $angle A$ and $angle F$ is $180^circ$ so quadrialteral $ABCD$ is cyclic. Because of that:
$$angle FAB=angle FCB=60^circ-frac x2=angle FDB$$
So triangle $FAD$ is isosceles and:
$$FA=FDtag{1}$$
In a similar way:
$$angle FAC=angle FBC=frac x2=angle FEA$$
So triangle $FEA$ is isosceles and:
$$FA=FEtag{2}$$
From (1) and (2) you have:
$$FD=FE$$
...and triangle $FDE$ must be isosceles. Angle $angle DFE=120^circ$ and therefore:
$$angle FDE=angle FED=30^circ$$
It follows that:
$$angle DEA=angle DEF+angle FEA=30^circ+frac x2$$
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
$endgroup$
1
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
add a comment |
$begingroup$
Drawing the rhombus is a nice way to start but we can finish differently.
As soon as you find that $triangle BDF$ is equilateral, you know that
$FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$
Referring to this same circle, since the central angle $angle BFD$ is $60^circ,$
the inscribed angle $angle BED$ is $30^circ.$
We have $angle CEF = angle ACB = x$ and $EB$ is the angle bisector of
$angle CEF,$ so $angle CEB = frac12 angle CEF = frac 12x.$
Then $angle CED = angle CEB + angle BED = frac 12x + 30^circ.$
answered Jan 22 at 18:04
David KDavid K
55k344120
$endgroup$
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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$begingroup$
$$angle CBA=120^circ-x$$
Triangle $DBC$ is isosceles:
$$angle CDB=angle DCB=frac12angle CBA=60^circ-frac x2$$
Triangle $BCE$ is isosceles:
$$angle CEB=angle CBE=frac12angle BCA=frac x2$$
Now, from triangle:
$$angle BFC=120^circ$$
Take a look at quadrilateral $ABFC$: The sum of opposite angles $angle A$ and $angle F$ is $180^circ$ so quadrialteral $ABCD$ is cyclic. Because of that:
$$angle FAB=angle FCB=60^circ-frac x2=angle FDB$$
So triangle $FAD$ is isosceles and:
$$FA=FDtag{1}$$
In a similar way:
$$angle FAC=angle FBC=frac x2=angle FEA$$
So triangle $FEA$ is isosceles and:
$$FA=FEtag{2}$$
From (1) and (2) you have:
$$FD=FE$$
...and triangle $FDE$ must be isosceles. Angle $angle DFE=120^circ$ and therefore:
$$angle FDE=angle FED=30^circ$$
It follows that:
$$angle DEA=angle DEF+angle FEA=30^circ+frac x2$$
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
$endgroup$
1
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
add a comment |
$begingroup$
$$angle CBA=120^circ-x$$
Triangle $DBC$ is isosceles:
$$angle CDB=angle DCB=frac12angle CBA=60^circ-frac x2$$
Triangle $BCE$ is isosceles:
$$angle CEB=angle CBE=frac12angle BCA=frac x2$$
Now, from triangle:
$$angle BFC=120^circ$$
Take a look at quadrilateral $ABFC$: The sum of opposite angles $angle A$ and $angle F$ is $180^circ$ so quadrialteral $ABCD$ is cyclic. Because of that:
$$angle FAB=angle FCB=60^circ-frac x2=angle FDB$$
So triangle $FAD$ is isosceles and:
$$FA=FDtag{1}$$
In a similar way:
$$angle FAC=angle FBC=frac x2=angle FEA$$
So triangle $FEA$ is isosceles and:
$$FA=FEtag{2}$$
From (1) and (2) you have:
$$FD=FE$$
...and triangle $FDE$ must be isosceles. Angle $angle DFE=120^circ$ and therefore:
$$angle FDE=angle FED=30^circ$$
It follows that:
$$angle DEA=angle DEF+angle FEA=30^circ+frac x2$$
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
$endgroup$
1
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
add a comment |
$begingroup$
$$angle CBA=120^circ-x$$
Triangle $DBC$ is isosceles:
$$angle CDB=angle DCB=frac12angle CBA=60^circ-frac x2$$
Triangle $BCE$ is isosceles:
$$angle CEB=angle CBE=frac12angle BCA=frac x2$$
Now, from triangle:
$$angle BFC=120^circ$$
Take a look at quadrilateral $ABFC$: The sum of opposite angles $angle A$ and $angle F$ is $180^circ$ so quadrialteral $ABCD$ is cyclic. Because of that:
$$angle FAB=angle FCB=60^circ-frac x2=angle FDB$$
So triangle $FAD$ is isosceles and:
$$FA=FDtag{1}$$
In a similar way:
$$angle FAC=angle FBC=frac x2=angle FEA$$
So triangle $FEA$ is isosceles and:
$$FA=FEtag{2}$$
From (1) and (2) you have:
$$FD=FE$$
...and triangle $FDE$ must be isosceles. Angle $angle DFE=120^circ$ and therefore:
$$angle FDE=angle FED=30^circ$$
It follows that:
$$angle DEA=angle DEF+angle FEA=30^circ+frac x2$$
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
$endgroup$
$$angle CBA=120^circ-x$$
Triangle $DBC$ is isosceles:
$$angle CDB=angle DCB=frac12angle CBA=60^circ-frac x2$$
Triangle $BCE$ is isosceles:
$$angle CEB=angle CBE=frac12angle BCA=frac x2$$
Now, from triangle:
$$angle BFC=120^circ$$
Take a look at quadrilateral $ABFC$: The sum of opposite angles $angle A$ and $angle F$ is $180^circ$ so quadrialteral $ABCD$ is cyclic. Because of that:
$$angle FAB=angle FCB=60^circ-frac x2=angle FDB$$
So triangle $FAD$ is isosceles and:
$$FA=FDtag{1}$$
In a similar way:
$$angle FAC=angle FBC=frac x2=angle FEA$$
So triangle $FEA$ is isosceles and:
$$FA=FEtag{2}$$
From (1) and (2) you have:
$$FD=FE$$
...and triangle $FDE$ must be isosceles. Angle $angle DFE=120^circ$ and therefore:
$$angle FDE=angle FED=30^circ$$
It follows that:
$$angle DEA=angle DEF+angle FEA=30^circ+frac x2$$
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
edited Jan 22 at 14:51
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
answered Jan 22 at 14:44
OldboyOldboy
8,62711036
8,62711036
1
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
add a comment |
1
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
1
1
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
In the very last line you mean $frac{x}{2}$.
$endgroup$
– rogerl
Jan 22 at 14:50
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@rogerl Thanks, I have corrected that.
$endgroup$
– Oldboy
Jan 22 at 14:51
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
$begingroup$
@Oldboy Wow. very nice. I learned a lot from your solution.
$endgroup$
– mina_world
Jan 22 at 14:58
add a comment |
$begingroup$
Drawing the rhombus is a nice way to start but we can finish differently.
As soon as you find that $triangle BDF$ is equilateral, you know that
$FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$
Referring to this same circle, since the central angle $angle BFD$ is $60^circ,$
the inscribed angle $angle BED$ is $30^circ.$
We have $angle CEF = angle ACB = x$ and $EB$ is the angle bisector of
$angle CEF,$ so $angle CEB = frac12 angle CEF = frac 12x.$
Then $angle CED = angle CEB + angle BED = frac 12x + 30^circ.$
answered Jan 22 at 18:04
David KDavid K
55k344120
$endgroup$
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
add a comment |
$begingroup$
Drawing the rhombus is a nice way to start but we can finish differently.
As soon as you find that $triangle BDF$ is equilateral, you know that
$FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$
Referring to this same circle, since the central angle $angle BFD$ is $60^circ,$
the inscribed angle $angle BED$ is $30^circ.$
We have $angle CEF = angle ACB = x$ and $EB$ is the angle bisector of
$angle CEF,$ so $angle CEB = frac12 angle CEF = frac 12x.$
Then $angle CED = angle CEB + angle BED = frac 12x + 30^circ.$
answered Jan 22 at 18:04
David KDavid K
55k344120
$endgroup$
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
add a comment |
$begingroup$
Drawing the rhombus is a nice way to start but we can finish differently.
As soon as you find that $triangle BDF$ is equilateral, you know that
$FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$
Referring to this same circle, since the central angle $angle BFD$ is $60^circ,$
the inscribed angle $angle BED$ is $30^circ.$
We have $angle CEF = angle ACB = x$ and $EB$ is the angle bisector of
$angle CEF,$ so $angle CEB = frac12 angle CEF = frac 12x.$
Then $angle CED = angle CEB + angle BED = frac 12x + 30^circ.$
answered Jan 22 at 18:04
David KDavid K
55k344120
$endgroup$
Drawing the rhombus is a nice way to start but we can finish differently.
As soon as you find that $triangle BDF$ is equilateral, you know that
$FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$
Referring to this same circle, since the central angle $angle BFD$ is $60^circ,$
the inscribed angle $angle BED$ is $30^circ.$
We have $angle CEF = angle ACB = x$ and $EB$ is the angle bisector of
$angle CEF,$ so $angle CEB = frac12 angle CEF = frac 12x.$
Then $angle CED = angle CEB + angle BED = frac 12x + 30^circ.$
answered Jan 22 at 18:04
David KDavid K
55k344120
answered Jan 22 at 18:04
David KDavid K
55k344120
answered Jan 22 at 18:04
David KDavid K
55k344120
answered Jan 22 at 18:04
David KDavid K
55k344120
55k344120
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
add a comment |
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
$begingroup$
Yes, good idea. This is a clean finish.
$endgroup$
– mina_world
Jan 22 at 18:34
add a comment |
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1
$begingroup$
If you want to see a solution that is different from the image you provided, it is better to describe what direction or type of approach you'd like to see. Basically if you don't show your own thoughts and efforts, the post will most likely be ignored or closed by community votes.
$endgroup$
– Lee David Chung Lin
Jan 22 at 12:29