Answer:
Step-by-step explanation:
Remark
There are many ways you can do this. You could, for example, prove this by using just one diagram and do the algebra for that one diagram. Or you could use both diagrams and equate them. I'll do the latter.
Left Diagram
c^2 + the 4 small triangles surrounding c^2 gives
c^2 + 4 * (1/2 * a * b)
Right Diagram
2 * a * b + a^2 + b^2
By the symmetry of the situation, you can find the area of the figures in each diagram. Since a in the left is equal to a on the right, and since b on the left = b on the right, the areas are equal.
Solution
c^2 + 4(1/2 a*b) = 2 * a * b + a^2 + b^2
c^2 + 2*a*b = 2* a * b + a^2 + b^2
c^2 = a^2 + b^2
Notice that the 2ab on the left cancels out with the 2ab on the right.
