Theorem 15.1.2

Theorem 15.1.2: In any triangle, the square on the side opposite to an acute angle is equal to sum of the squares on the sides containing that acute angle diminished by twice the rectangle contained by one of the sides and the projection on it of the other.

Given: $\triangle ABC$ with an acute $\angle CAB$ at vertex A.

Taking, $m\overline{BC} = a, m\overline{AC} = b$ and $m\overline{AB} = c$.

Construction: Draw $\overline{CD}\; ꓕ\; \overline{AC}$ so that $\overline{AD}$ is projection of $\overline{AC}$ on $\overline{AB}$.

Also suppose, $m\overline{AD}$ = x and $m\overline{CD} = h$

To prove: $(m\overline{BC})^2 = (m\overline{AC})^2 + (m\overline{AB})^2 - 2(m\overline{AB})(m\overline{AD})$ i.e. $a^2 = b^2 + c^2 - 2cx$

Proof: