Find all such triangles ABC such that $AB+AC =2$cm and $AD+BC = \sqrt{5}$ cm where AD is the altitude through A. Find all such triangles ABC such that $AB+AC =2$cm and $AD+BC = \sqrt{5}$ cm 
where AD is the altitude through A.
I got 3 equations but there are 4 variables. So, its not working. Maybe  sine rule can work
 A: Let $BC=a,CA=b,AB=c$. Also, let $S$ be the area of $\triangle{ABC}$. Then, having that 
$$S=\frac 12\times a\times AD\quad\Rightarrow\quad AD=\frac{2S}{a}=\frac{bc\sin\theta}{a}$$ where $\theta=\angle{BAC}$,
we have
$$c+b=2$$
$$\frac{bc\sin\theta}{a}+a=\sqrt 5$$
Then,
$$\sin\theta=\frac{a(\sqrt 5-a)}{b(2-b)}$$
By the law of cosines,
$$a^2=b^2+(2-b)^2-2b(2-b)\cos\theta\Rightarrow \cos\theta=\frac{b^2+(2-b)^2-a^2}{2b(2-b)}$$
Now, from $\cos^2\theta+\sin^2\theta=1$, we have
$$\left(\frac{b^2+(2-b)^2-a^2}{2b(2-b)}\right)^2+\left(\frac{a(\sqrt 5-a)}{b(2-b)}\right)^2=1,$$
i.e.
$$4(4-a^2)b^2+8(a^2-4)b+5a^4-8\sqrt 5\ a^3+12a^2+16=0$$
and so we have to have
$$(8(a^2-4))^2-4\cdot 4(4-a^2)(5a^4-8\sqrt 5\ a^3+12a^2+16)\ge 0$$$$\iff  (a-2)(a+2) (\sqrt 5\ a-4)^2\ge 0\iff a=\frac{4}{\sqrt 5}\quad\text{or}\quad a\ge 2$$
Also, we have to have
$$a\lt b+c\iff a\lt 2.$$
Thus, $BC=a$ has to be $\frac{4}{\sqrt 5}$, from which we have $AC=AB=1$. This is sufficient.
A: 
Let $BC=a$, then $h_a=\sqrt5-a$. On a straight line parallel $BC$, the minimum value of the sum of the distances from its point $A$ to the heights $B$ and $C$ observed in the case of an isosceles triangle. The value of this sum is equal to $2\sqrt{(\frac{a}2)^2+(\sqrt5-a)^2}$, and it is no longer $2$. Here $\frac{a^2}4+a^2+5-2a\sqrt5\le1$, i.e $\frac54a^2-2a\sqrt5+4\le0$. The discriminant is zero, and inequality has exactly one root $a=\frac45\sqrt5$. The triangle is isosceles with a side of $1$.
