Since AB=Bc, triangle ABC is isosceles with base AC. Then BN is height and median drawn to the base. If BN is median and AC=12 cm, then AN=NC=6 cm.
1. By the Pythagorean theorem for triangle ABN,
[tex]AB^2=AN^2+BN^2,\\ \\10^2=6^2+BN^2,\\ \\BN^2=100-36=64,\\ \\BN=8\ cm.[/tex]
2. AB and BC are legs of isosceles triangle ABC. The heights drawn to the lags of isosceles triangle are congruent, so AK=CM. Consider two right triangles AKC and AKB. By the Pythagorean theorem,
[tex]AC^2=AK^2+KC^2,\\ \\AB^2=BK^2+AK^2=(BC-KC)^2+AK^2.[/tex]
Subtract from the second equation the first one:
[tex]AB^2-AC^2=(BK-KC)^2-KC^2,\\ \\10^2-12^2=(10-KC)^2-KC^2,\\ \\100-144=100-20KC+KC^2-KC^2,\\ \\-144=-20KC,\\ \\KC=7,2\ cm.[/tex]
Then
[tex]AK^2=AC^2-KC^2,\\ \\AK^2=12^2-7.2^2=144-51,84=92.16,\\ \\AK=9.6\ cm.[/tex]
Answer: AK=CM=9.6 cm, BN=8 cm
