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|NextLevel = [[Perpendicular Bisector]] |
|NextLevel = [[Perpendicular Bisector]] |
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− | == |
+ | ==Solutions== |
− | ===3L 3E=== |
+ | ===3L 3E Solution=== |
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Let A be the initial point of the given ray. |
Let A be the initial point of the given ray. |
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#Construct the circle C2 with center B and radius AB, intersecting circle C1 at C and D |
#Construct the circle C2 with center B and radius AB, intersecting circle C1 at C and D |
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#Construct line AC |
#Construct line AC |
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+ | ---- |
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− | ===2V=== |
+ | ===2V Solution=== |
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#Construct line AD |
#Construct line AD |
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==Explanation== |
==Explanation== |
Revision as of 23:17, 2 December 2020
Solutions
3L 3E Solution
Let A be the initial point of the given ray.
- Construct a circle C1 with center A and an arbitrary radius, intersecting the ray at B
- Construct the circle C2 with center B and radius AB, intersecting circle C1 at C and D
- Construct line AC
2V Solution
- Construct line AD
Explanation
All internal angles of an equilateral triangle are 60°, so you just need to construct the two sides of an equilateral triangle adjacent to vertex A.
The two circles ensure that AB=AC=BC and AB=AD=BD and so ABC and ABD are equilateral and angles CAB and DAB are 60°.