Euclid Geometry: Seeking for a simpler geometric solution The Next CEO of Stack OverflowCongruent parts of trianglesFound a New Golden Ratio Construction with Equilateral Triangle, Square, and Circle. Geometric/Trigonmetric proof?Proof in Geometry. Is this solution correct?What is a geometric construction corresponding to elliptic curve addition for Sharygin-isosceles triangles?Problem about angle in isosceles triangleInscribing rhombus in a triangle's angle in only eight compass-and-straightedge stepsIn the figure, prove that points D, M and P are collinear.Angle in a 80-80-20 triangleIs there a proof of this question through Napoleon's theorem?Find an angle in a geometric figure (given) considering triangles
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Euclid Geometry: Seeking for a simpler geometric solution
The Next CEO of Stack OverflowCongruent parts of trianglesFound a New Golden Ratio Construction with Equilateral Triangle, Square, and Circle. Geometric/Trigonmetric proof?Proof in Geometry. Is this solution correct?What is a geometric construction corresponding to elliptic curve addition for Sharygin-isosceles triangles?Problem about angle in isosceles triangleInscribing rhombus in a triangle's angle in only eight compass-and-straightedge stepsIn the figure, prove that points D, M and P are collinear.Angle in a 80-80-20 triangleIs there a proof of this question through Napoleon's theorem?Find an angle in a geometric figure (given) considering triangles
$begingroup$
The problem:
Answer:
Extend $AC$ to the side of $C$, such that $BC=CE$. Then, construct the
equilateral triangle $BEF$. Triangles $CAB$ and $EFC$ are congruent,
thus $AC=BE$. Now, observe that $AC=DE$, thus $BE=DE$. It follows that
$triangle BED$ is isosceles, thus $angle EDB=80$.
I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?
euclidean-geometry
asked Mar 28 at 9:54
blackenedblackened
485413
$endgroup$
add a comment |
$begingroup$
The problem:
Answer:
Extend $AC$ to the side of $C$, such that $BC=CE$. Then, construct the
equilateral triangle $BEF$. Triangles $CAB$ and $EFC$ are congruent,
thus $AC=BE$. Now, observe that $AC=DE$, thus $BE=DE$. It follows that
$triangle BED$ is isosceles, thus $angle EDB=80$.
I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?
euclidean-geometry
asked Mar 28 at 9:54
blackenedblackened
485413
$endgroup$
add a comment |
$begingroup$
The problem:
Answer:
Extend $AC$ to the side of $C$, such that $BC=CE$. Then, construct the
equilateral triangle $BEF$. Triangles $CAB$ and $EFC$ are congruent,
thus $AC=BE$. Now, observe that $AC=DE$, thus $BE=DE$. It follows that
$triangle BED$ is isosceles, thus $angle EDB=80$.
I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?
euclidean-geometry
asked Mar 28 at 9:54
blackenedblackened
485413
$endgroup$
The problem:
Answer:
Extend $AC$ to the side of $C$, such that $BC=CE$. Then, construct the
equilateral triangle $BEF$. Triangles $CAB$ and $EFC$ are congruent,
thus $AC=BE$. Now, observe that $AC=DE$, thus $BE=DE$. It follows that
$triangle BED$ is isosceles, thus $angle EDB=80$.
I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?
euclidean-geometry
euclidean-geometry
asked Mar 28 at 9:54
blackenedblackened
485413
asked Mar 28 at 9:54
blackenedblackened
485413
asked Mar 28 at 9:54
blackenedblackened
485413
asked Mar 28 at 9:54
blackenedblackened
485413
asked Mar 28 at 9:54
blackenedblackened
485413
485413
add a comment |
add a comment |
0
active
oldest
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