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

[Albert1]

$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

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

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

 

[RLBrown]

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

 

[Albert1]

$\triangle ABC, \,\, AB=AC$,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

 

[Albert1]

$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

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

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

Spoiler

View attachment 2741