Giant triangle!

Let an isosceles triangle have vertices with the following coordinates

$(-s,0)$
$(0,0)$
$(\sqrt{{s}^{2}-{0.3}^{2}}, 0.3)$

then the sum of the altitudes will be $< 1$ , while the area of the triangle will be $0.15s$ , and $s$ can be as large as one wants.

Let consider an isosceles triangle $ABC$ such that $AB=b=AC=c$ , $BC=a=6\times { 10 }^{ 14 }km$ , the height from $A$ ${ h }_{ a }=0.2mm=2\times { 10 }^{ -7 }km$ . ( $A$ can be formed from the bisector of the given line $BC$ and the given distance from $A$ to $BC$ )

We can see that $\widehat { BAC } =\alpha$ is very very near to ${ 180 }^{ 0 }$ ( ${ 179.999... }^{ 0 }$ ) and the area of $ABC$ is $6\times { 10 }^{ 7 }{ km }^{ 2 }\quad \left( >5.1\times { 10 }^{ 7 } \right)$

Since $b=c>\frac { a }{ 2 }$ (Triangle inequality) $\Longrightarrow { h }_{ b }={ h }_{ c }<2{ h }_{ a }\Longrightarrow { { h }_{ a }+h }_{ b }+{ h }_{ c }<5{ h }_{ a }=1mm$ .

Therefore, the triangle $ABC$ above satisfies all the conditions $\Longrightarrow$ The answer is $\boxed { True }$ .