Many thank you for your comments I must apologize for the errors and for being too quickie, below is a more detailed exposition of my reasoning.

We can take on the plane two arbitrary points A and B provided that its distance between them being less than the sum a+b, so vertex C is on the ellipse focii A and B. By stretching or shortening the segment AB but keeping the condition to be less than a+b we get a family of ellipses.

Simultaneously since C is the vertex of the straight angle of our sought triangle must belong to a circumference diameter c, unknown. So by making c equal to the distance AB we have four possible locations in the ellipse for point C the intersections of both curves. Now we can narrow the distance between focii till the two upper and the two lower points coincide, in other words the circumference being tangent to the ellipse. Then the diameter of the circumference equals to the minor axis of the ellipse and to the focal distance, which means sides a ad b are equal, the altitude h = c/2 = a/sqrt(2), what I stated that wrong before.

Another even quicker approach is considering a rectangle formed with two triangles perimeter equal 2(a+b) and diagonals c, the biggest area for this rectangle which also render the biggest altitude is when diagonals are perpendicular which means our rectangle is a square.