Given a triangle $ABC$ with angle $A$ obtuse and with altitudes of length $h$ and $k$ as shown in the diagram, prove that $a+h\ge b+k$. Find under what conditions $a+h=b+k$.
[asy]
size(6cm);

pair A = dir(105), C = dir(170), B = dir(10), D = foot(B, A, C), E = foot(A, B, C);

draw(A--B--C--cycle);
draw(B--D--A--E);

dot(A); dot(B); dot(C); dot(D); dot(E);

label("$A$", A, dir(110));
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, D);
label("$E$", E, dir(45));

label("$h$", A--E, dir(0));
label("$k$", B--D, dir(45));
[/asy]
1. Given a triangle \(ABC\) with \(\angle A\) obtuse, we need to prove that \(a + h \ge b + k\), where \(h\) and \(k\) are the lengths of the altitudes from vertices \(A\) and \(B\) respectively.

2. First, express the altitudes \(h\) and \(k\) in terms of the sides and angles of the triangle:
   \[
   h = b \sin C \quad \text{and} \quad k = a \sin C
   \]
   This follows from the definition of the altitude in a triangle, where the altitude is the perpendicular distance from a vertex to the opposite side.

3. Substitute these expressions into the inequality \(a + h \ge b + k\):
   \[
   a + b \sin C \ge b + a \sin C
   \]

4. Rearrange the terms to isolate the sine terms on one side:
   \[
   a - a \sin C \ge b - b \sin C
   \]

5. Factor out the common terms:
   \[
   a(1 - \sin C) \ge b(1 - \sin C)
   \]

6. Since \(\sin C \le 1\) for any angle \(C\), and \(\angle A\) is obtuse, implying \(\angle C\) is acute or right, we have \(1 - \sin C \ge 0\). Therefore, the inequality holds if and only if:
   \[
   a \ge b
   \]
   This is true because \(\angle A\) is the largest angle in the triangle, making \(a\) the longest side.

7. Equality occurs when:
   \[
   (1 - \sin C)(a - b) = 0
   \]
   This can happen in two cases:
   - \(\sin C = 1\), which implies \(\angle C = 90^\circ\). However, this contradicts the obtuseness of \(\angle A\).
   - \(a = b\), which implies the triangle is isosceles with \(A\) being obtuse.

Therefore, the condition for equality \(a + h = b + k\) is when the triangle is isosceles with \(a = b\) and \(\angle A\) obtuse.

\(\blacksquare\)

The final answer is \( \boxed{ a + h \ge b + k } \) with equality when \(a = b\) and \(\angle A\) is obtuse.