Segments

Given a triangle ABC (with BC/2 < AC < 2BC, why ?)
Find D on BC and E on AC with AE = DE = BD, using a collapsing compass and (unmarked) straightedge.
Hint

Solution
Animated JavaSketchpad (wait for initializing and loading of Applet)

We'll care now of triangles ABC for which DE is perpendicular to BC.
Locus of D, E, C for a given AB.

Solution

 

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