The angle between the lateral sides of an isosceles triangle is less than 60, the median and the height are drawn

The angle between the lateral sides of an isosceles triangle is less than 60, the median and the height are drawn to the lateral side, the lengths of which are 3 * √ (5) and 6. Find the length of the lateral side.

Triangle ABC: AB = BC, AM = 3√5 – median, AH = 6 – height.
1. From the triangle АHM (by the Pythagorean theorem):
МН = √ (AM ^ 2 – AH ^ 2) = √ ((3√5) ^ 2 – 6 ^ 2) = √ (9 * 5 – 36) = √ (45 – 36) = √9 = 3.
2. Since AM is the median then:
BM = MC.
Point H divides the MC into segments:
MC = MH + CH.
Then:
MB = MH + CH.
Substitute the value of MH:
BM = MC = 3 + CH.
3. BC consists of segments:
BC = BM + C = 2 (3 + CH) = 6 + 2CH.
Since AB = BC, then:
AB = 6 + 2CH.
4. From the triangle AНВ (by the Pythagorean theorem):
AB = √ (BH ^ 2 + AH ^ 2);
6 + 2CH = √ ((BM + MH) ^ 2 + 6 ^ 2);
6 + 2CH = √ ((3 + CH + 3) ^ 2 + 6 ^ 2);
6 + 2CH = √ ((6 + CH) ^ 2 + 6 ^ 2);
6 + 2CH = √ (36 + 12CH + CH ^ 2 + 36);
6 + 2CH = √ (CH ^ 2 + 12CH + 72) (square both sides of the equation);
36 + 24CH + 4CH ^ 2 = CH ^ 2 + 12CH + 72;
4CH ^ 2 – CH ^ 2 + 24CH – 12CH + 36 – 72 = 0;
3СН ^ 2 + 12СН – 36 = 0;
CH ^ 2 + 4CH – 12 = 0.
Discriminant:
D = 16 + 48 = 64.
CH1 = (- 4 + 8) / 2 = 4/2 = 2.
CH2 = (- 4 – 8) / 2 = -12/2 = -6 – does not satisfy the meaning of the problem.
Thus, the CH segment is equal to 2.
5. Find the length of the MC:
MS = MH + CH = 3 + 2 = 5.
BC = 2MC = 2 * 5 = 10.
Then:
AB = BC = 10.
Answer: AB = BC = 10.