Let D be the point on BD=2DC. D is one possible location for M. Area(ABD)/area(ACD)=2
Now let M' be any point on line AD (except A)
area(ABM')/area(ABD) = AM'/AD = area(ACM')/area(ACD)
So area(ABM')/area(ACM')=2
If M' were not on line line AD the areas will change according to the ratio BD'/CD' where D' is the intersection of lines AM' and DC.
Finally a note: If we remove the restriction that M is inside triangle ABC, the locus includes a second line.
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Posted by Jer
on 2023-10-09 13:10:38 |