In an isosceles triangle with side a and height h drawn to the base, find the length of the vector that coincides with

In an isosceles triangle with side a and height h drawn to the base, find the length of the vector that coincides with the median drawn to the side.

Let △ ABC be given: AB = BC = a, BH = h – height.
1. From △ AHB: ∠AHB = 90 °, AH = AC / 2 (since BH is both height, median, and bisector) and BH = h – legs, AB = a – hypotenuse.
By the Pythagorean theorem:
AH = √ (AB² – BH²) = √ (a² – h²).
2. AM is the median drawn to the lateral side.
AM and BH meet at point O.
The medians are divided by the intersection point in a ratio of 2: 1 starting from the top, then:
BO / HO = 2/1;
BO = 2 * HO.
Also:
BO + HO = BH;
2 * HO + HO = h;
3 * HO = h;
HO = h / 3.
3. From △ AHO: ∠AHO = 90 °, AH = √ (a² – h²) and HO = h / 3 – legs, AO – hypotenuse.
By the Pythagorean theorem:
AO = √ (AH² + HO²) = √ ((√ (a² – h²)) ² + (h / 3) ²) = √ (a² – h² + h² / 9) = √ ((9 * a² – 9 * h² + h²) / 9) = (√ (9 * a² – 8 * h²)) / 3.
4. Since the medians are divided by the intersection point in a ratio of 2: 1 starting from the top, then:
AO / MO = 2/1;
MO = AO / 2;
MO = (√ (9 * a² – 8 * h²)) / 3: 2 = (√ (9 * a² – 8 * h²)) / 3 * ½ = (√ (9 * a² – 8 * h²)) / 6 …
5. The length of the median AM is:
AM = AO + MO = (√ (9 * a² – 8 * h²)) / 3 + (√ (9 * a² – 8 * h²)) / 6 = (2 * √ (9 * a² – 8 * h²) + √ (9 * a² – 8 * h²)) / 6 = (3 * √ (9 * a² – 8 * h²)) / 6 = √ (9 * a² – 8 * h²) / 2.
Answer: AM = √ (9 * a² – 8 * h²) / 2.