ABC is an isosceles triangle. AK is the bisector. AB = BC = 20. The base is 5. Find AK.

Consider an isosceles triangle ABC, where AB = BC = 20, AC = 5,

AK is the bisector of the angle BAC.

By the property of the bisector, we have:

BK / KC = AB / AC = 20/5 = 4,

BK = 4 * KC.

Then, since:

BC = BK + KC = 4 * KC + KC = 5 * KC = 20,

KC = 4.

Let us lower the height from the top BH to the base of the AC.

Since triangle ABC is isosceles, AH = CH = 1/2 * AC = 1/2 * 5 = 5/2.

The BHC triangle is rectangular. Therefore, we have:

CH = BC * cos (C),

5/2 = 20 * cos (C),

cos (C) = 5/40.

From the triangle AKC by the cosine theorem we get:

AK ^ 2 = AC ^ 2 + KC ^ 2 – 2 * AC * KC * cos (C) = 5 ^ 2 + 4 ^ 2 – 2 * 5 * 4 * 5/40 = 25 + 16 – 5 = 36,

AK = 6.

Answer: the length of the bisector AK = 6.