The lateral side of an isosceles triangle is 9 cm less than the base, and the segments into which the bisector

The lateral side of an isosceles triangle is 9 cm less than the base, and the segments into which the bisector at the base divides the height drawn to the base are 5: 4. Find the height of the triangle drawn to the base.

Let’s designate this triangle ABC, base AC = x cm, lateral side AB = (x – 9) cm. Height BH, point O – the point of intersection of the bisector and height.
Consider a triangle ABН and, by the property of the bisector of an angle, we write down the ratio of the sides of this triangle:
AB / AH = BO / OH = 5/4.
AH = x / 2 (half of the base, triangle ABC is isosceles by condition).
(x – 9) / x / 2 = 5/4
4x – 36 = 5x / 2
8x – 72 = 5x
x = 24 (cm) – the base of the speaker.
AB = 24 – 9 = 15 (cm).
AH = 24/2 = 12 (cm).
According to the Pythagorean theorem, in the ABН triangle, we find the BN leg (the height of the ABC triangle):
BH = √ (AB² – AH²) = √ (225 – 144) = √81 = 9 (cm).
Answer: 9 cm is the height of the triangle.