Construct, with straightedge and compass, a ΔABM given |AB| and |BM| such that the altitude from vertex A, the median from vertex M, and the bisector of ∠ABM are concurrent.
This is quite easy for given |AB| and ∠ABM.
Bisect AB.
Bisect ∠ABM.
Construct the perpendicular from A to BM.
The point where the angle bisector and altitude cross is the point where the median must also cross. Construct the line through this point and the bisector of AB.
The third point (M) of the triangle is where this line intersects the other ray from ∠ABM.
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Posted by Jer
on 2011-05-03 14:59:03 |
with different givens
