Problem 24
Let ABC be an equilateral triangle. Let D be a point on the circumcircle of ABC, such that D is the midpoint of minor arc AB. Let E and F be points on AC and BC respectively. Then, show that
AE + FB = EF ⇔ ∠EDF = 60°
Here’s a simulation that demonstrates the problem.
Try dragging the point E. The point F will move on BC such that ∠EDF = 60°. Is AE + BF always equal to EF? Will the converse be also true? How can we prove the same?
Solution
Angle in a Semicircle
[coming soon]
Congruent Triangles
[coming soon]
Black Magic
[coming soon]
Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Telegram group.
Problem Source: CutTheKnot
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