Given any acute angled triangle $ABC$, left points $A',B',C'$ be located as follows : $A'$ is the point where altitude from A on $BC$ meets the v outwards facing semicircle drawn on BC as diameter. Pointed $B',C'$ are located similarly .Prove that : $[BCA']^2+[CAB']^2+[ABC']^2=[ABC]^2$. Where [•] denote area of triangles.
Above is the problem. We know that the area of a triangle is $\dfrac12 \cdot \text{height} \cdot \text{base}$. But I can't proceed.
Hint: Prove first the following generalization of the Pythagorean theorem:
then prove that in your configuration $A'B=C'B,B'A=C'A,A'C=B'C$, hence you may conclude from the previous lemma. How to prove that $A'B=C'B$? Simple. We have: $$ A'B^2+A'C^2 = a^2, \qquad A'B^2-A'C^2 = AB^2-AC^2 = c^2-b^2 $$ (since $BB'\perp AC$) hence $A'B^2=\frac{a^2+c^2-b^2}{2}=C'B^2$.
Like Desargues' theorem, this $2d$-problem is easier to prove if we consider it in $3d$!