MHB Max Area of $\triangle ABC$ with $AB=AC$ and $BD=m$

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In the discussion, the focus is on maximizing the area of triangle ABC, where AB equals AC and D is the midpoint of AC. The area n is expressed in terms of the distance BD, denoted as m. A specific case is analyzed with m set to 3/2, leading to calculations of y and x, ultimately revealing that the area approaches zero at extremes. The law of cosines is suggested as a method to derive a general solution for maximizing the area and determining the corresponding angle A. The goal is to find the maximum area n and the angle A that achieves this maximum.
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$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

if $BD=m$,and $n$=area of $\triangle ABC$

please find $max(n)$ and corresponding $\angle A$
 
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Here's one approach...

Parameters {x,y}:
C={x,0}, B={-x,0} A={0,y} implies D={x/2,y/2}

Set Scale: try m=3/2
Extremes:
then y=0 implies x=1 and n=0
then y=3 implies x=0 and n=0

get y(x) for m=3/2
Solve case m=3/2 and scale for general case
 
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Albert said:
$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

if $BD=m$,and $n$=area of $\triangle ABC$

please find $max(n)$ and corresponding $\angle A$
hint :use law of cosine
 
Albert said:
$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

if $BD=m$,and $n$=area of $\triangle ABC$

please find $max(n)$ and corresponding $\angle A$
 

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