Prove that in any acute angled \(\ ABC ) is \(c2 = a2 + b2 − 2ab C ).

Triangle Setup:

Imagine an acute-angled triangle $\Delta ABC$.

• Label the vertices A, B, and C.
• The side opposite vertex A is denoted by $a$.
• The side opposite vertex B is denoted by $b$.
• The side opposite vertex C is denoted by $c$.
• Draw an altitude (a perpendicular line) from vertex B to the side AC. Let the point where this altitude meets AC be D.
• This altitude BD divides $\Delta ABC$ into two right-angled triangles: $\Delta ADB$ and $\Delta CDB$.
• Let the length of the altitude BD be $h$.
• Let the length of the segment AD be $x$.
• Since AC is side $b$, the length of the segment DC will be $b - x$.
• Angle C is the angle at vertex C.

The detailed explanation and proof for $c^2 = a^2 + b^2 - 2ab \cos C$ using this triangle setup was provided in our previous conversation.