Derive Law of Cosines using Pythagorean Theorem

Consider the triangle ABC. Denote h the altitude through B and D the point where h intersects the (extended) base AC
Cosine function for triangle ADB.

cos α= x/c  =>  x=c*cos α
 

Pythagorean theorem for triangle ADB
x²+h²=c²*x²+h²=c²
h²=c²−x²*h²=c²−x²

Pythagorean theorem for triangle CDB
(b−x)²+h²=a²*(b−x)²+h²=a²

Substitute h² = c² - x²
(b−x)²+(c²−x²)=a²(b−x)²+(c²−x²)=a²
(b²−2bx+x²)+(c²−x²)=a²(b²−2bx+x²)+(c²−x²)=a²
b²−2bx+c²=a²b²−2bx+c²=a²

Substitute x = ccos α
b²−2b(ccosα)+c²=a²b²−2b(c*cos α)+c²=a²

Rearrange to get Law of Cosines

a²=b²+c²−2bc*cos α