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 Question from Nazrul, a teacher: How can I draw an isosceles triangle whose each angle adjacent to the base is twice the vertex angle? Please help me.

Nazrul,

If you do the arithmetic, the problem is to construct a 36 degree angle -- that is the first step to constructing a regular pentagon. Euclid did this in Book IV, Proposition 11. You can find his method and many others on the internet. Here's my favorite method: First inscribe a regular hexagon in a circle (which means drawing six points on a circle with the distance between successive points equal to the radius of the circle). Join alternate points to form an equilateral triangle ABC. Let D be the midpoint of AB and E of AC, and call G the point closest to F where EF meets the circumcircle. Then FG is the side of a regular pentagon whose diagonal has length EF. You use FG and EF to draw your 72 degree angle. (The reason this works is that the ratio of EF to FG is the golden section: $EF^2 = EF + FG.$)

Chris

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